Linear categories in Lawvere's Conceptual Mathematics
$begingroup$
In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
And then makes the following claim:
Map identity with f,g and h
I've tried in vain to prove Lawvere's claim (following his definitions) that the product of these two matrices must indeed have the form that he stated, with the first row being $1,h$ and the second row being $0,1$. No additional explanation is provided in the text.
Why must this be true?
category-theory
asked Sep 9 '18 at 10:57
user97678user97678
976
$endgroup$
add a comment |
$begingroup$
In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
And then makes the following claim:
Map identity with f,g and h
I've tried in vain to prove Lawvere's claim (following his definitions) that the product of these two matrices must indeed have the form that he stated, with the first row being $1,h$ and the second row being $0,1$. No additional explanation is provided in the text.
Why must this be true?
category-theory
asked Sep 9 '18 at 10:57
user97678user97678
976
$endgroup$
1
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Sep 9 '18 at 11:12
add a comment |
$begingroup$
In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
And then makes the following claim:
Map identity with f,g and h
I've tried in vain to prove Lawvere's claim (following his definitions) that the product of these two matrices must indeed have the form that he stated, with the first row being $1,h$ and the second row being $0,1$. No additional explanation is provided in the text.
Why must this be true?
category-theory
asked Sep 9 '18 at 10:57
user97678user97678
976
$endgroup$
In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
And then makes the following claim:
Map identity with f,g and h
I've tried in vain to prove Lawvere's claim (following his definitions) that the product of these two matrices must indeed have the form that he stated, with the first row being $1,h$ and the second row being $0,1$. No additional explanation is provided in the text.
Why must this be true?
category-theory
category-theory
asked Sep 9 '18 at 10:57
user97678user97678
976
asked Sep 9 '18 at 10:57
user97678user97678
976
asked Sep 9 '18 at 10:57
user97678user97678
976
asked Sep 9 '18 at 10:57
user97678user97678
976
asked Sep 9 '18 at 10:57
user97678user97678
976
976
1
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Sep 9 '18 at 11:12
add a comment |
1
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Sep 9 '18 at 11:12
1
1
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Sep 9 '18 at 11:12
$begingroup$
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
$endgroup$
– José Carlos Santos
Sep 9 '18 at 11:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The product $Xtimes Y$ implicitly comes with two fixed projections $pi_X,pi_Y$, and these play an implicit role in the definitions.
Dually, the coproduct $X+Y$ comes equipped with inclusions $iota_X,iota_Y$.
First, observe that (writing composition to the right),
$$underset{X+Y,to, Xtimes Y,to, X}{pmatrix{a&b\c&d}pi_X} = pmatrix{a\c}$$
and similarly, composing it by $pi_Y$ yields the second column.
Consequently, as $1_{Xtimes Y}=alpha,pmatrix{1&0\0&1}$, we have
$$ pi_X = alpha,pmatrix{1&0\0&1}pi_X =underset{Xtimes Yto X+Yto X}{qquad alpha,pmatrix{1\0}},. $$
Dual statements hold for rows and the $iota$'s.
From here, we can evaluate the 3 required elements like:
$$iota_Ycdot,pmatrix{1&f\0&1}alphapmatrix{1&g\0&1},cdotpi_X =
pmatrix{0&1},alpha,pmatrix{1\0} = pmatrix{0&1}pi_X = 0,.$$
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
$endgroup$
1
$begingroup$
I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
$endgroup$
– user97678
Sep 10 '18 at 7:36
1
$begingroup$
Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
$endgroup$
– Berci
Sep 10 '18 at 7:46
1
$begingroup$
I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
$endgroup$
– user97678
Sep 10 '18 at 8:20
1
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
$endgroup$
– user97678
Sep 10 '18 at 17:52
1
$begingroup$
Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
$endgroup$
– Berci
Sep 10 '18 at 21:38
|
show 3 more comments
$begingroup$
This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A times B overset{alpha_{AB}}{to} A +B $$ denotes the inverse of the "identity matrix"
$$begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} $$
then we have the following identities:
$$ alpha_{AB} circ langle 1_A, 0_{AB} rangle = j_{A+B}^1 ,, quad alpha_{AB} circ langle 0_{BA}, 1_B rangle = j_{A+B}^2 ,, quad begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} = pi_{A times B}^1 ,, quad begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} = pi_{A times B}^2 ,.$$
The proofs follow from the definition of $alpha_{AB}$, namely:
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^1 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = alpha_{AB} circ langle 1_A, 0_{AB} rangle $$
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^2 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = alpha_{AB} circ langle 0_{BA} , 1_B rangle $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^1 = left( pi_{A times B}^1 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^1 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^1 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^2 = left( pi_{A times B}^2 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^2 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^2 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} $$ $triangle$
With that under our belt, the result is fairly straightforward:
$$ begin{bmatrix} 1_A & f \ 0_{BA} & 1_B end{bmatrix} cdot begin{bmatrix} 1_A & g \ 0_{BA} & 1_B end{bmatrix} = begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} circ begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} =: begin{bmatrix} h_{AA} & h_{AB} \ h_{BA} & h_{BB} end{bmatrix} ,.$$
Therefore we have that:
$$ h_{AA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 1_A, f rangle = pi_{Atimes B}^1 circ langle 1_A, f rangle = 1_A $$
$$ h_{BA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle =begin{array}{c} pi_{Atimes B}^1 circ langle 0_{BA}, 1_B rangle \ begin{cases}1_A \ 0_{BA} end{cases} circ j_{A+B}^2 end{array} = 0_{BA} $$
$$ h_{BB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} g \ 1_{B} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle = begin{cases} g \ 1_{B} end{cases} circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} g \ 1_B end{cases} circ alpha_{AB} circ langle 1_A, f rangle ,. $$
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
The product $Xtimes Y$ implicitly comes with two fixed projections $pi_X,pi_Y$, and these play an implicit role in the definitions.
Dually, the coproduct $X+Y$ comes equipped with inclusions $iota_X,iota_Y$.
First, observe that (writing composition to the right),
$$underset{X+Y,to, Xtimes Y,to, X}{pmatrix{a&b\c&d}pi_X} = pmatrix{a\c}$$
and similarly, composing it by $pi_Y$ yields the second column.
Consequently, as $1_{Xtimes Y}=alpha,pmatrix{1&0\0&1}$, we have
$$ pi_X = alpha,pmatrix{1&0\0&1}pi_X =underset{Xtimes Yto X+Yto X}{qquad alpha,pmatrix{1\0}},. $$
Dual statements hold for rows and the $iota$'s.
From here, we can evaluate the 3 required elements like:
$$iota_Ycdot,pmatrix{1&f\0&1}alphapmatrix{1&g\0&1},cdotpi_X =
pmatrix{0&1},alpha,pmatrix{1\0} = pmatrix{0&1}pi_X = 0,.$$
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
$endgroup$
1
$begingroup$
I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
$endgroup$
– user97678
Sep 10 '18 at 7:36
1
$begingroup$
Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
$endgroup$
– Berci
Sep 10 '18 at 7:46
1
$begingroup$
I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
$endgroup$
– user97678
Sep 10 '18 at 8:20
1
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
$endgroup$
– user97678
Sep 10 '18 at 17:52
1
$begingroup$
Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
$endgroup$
– Berci
Sep 10 '18 at 21:38
|
show 3 more comments
$begingroup$
The product $Xtimes Y$ implicitly comes with two fixed projections $pi_X,pi_Y$, and these play an implicit role in the definitions.
Dually, the coproduct $X+Y$ comes equipped with inclusions $iota_X,iota_Y$.
First, observe that (writing composition to the right),
$$underset{X+Y,to, Xtimes Y,to, X}{pmatrix{a&b\c&d}pi_X} = pmatrix{a\c}$$
and similarly, composing it by $pi_Y$ yields the second column.
Consequently, as $1_{Xtimes Y}=alpha,pmatrix{1&0\0&1}$, we have
$$ pi_X = alpha,pmatrix{1&0\0&1}pi_X =underset{Xtimes Yto X+Yto X}{qquad alpha,pmatrix{1\0}},. $$
Dual statements hold for rows and the $iota$'s.
From here, we can evaluate the 3 required elements like:
$$iota_Ycdot,pmatrix{1&f\0&1}alphapmatrix{1&g\0&1},cdotpi_X =
pmatrix{0&1},alpha,pmatrix{1\0} = pmatrix{0&1}pi_X = 0,.$$
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
$endgroup$
1
$begingroup$
I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
$endgroup$
– user97678
Sep 10 '18 at 7:36
1
$begingroup$
Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
$endgroup$
– Berci
Sep 10 '18 at 7:46
1
$begingroup$
I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
$endgroup$
– user97678
Sep 10 '18 at 8:20
1
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
$endgroup$
– user97678
Sep 10 '18 at 17:52
1
$begingroup$
Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
$endgroup$
– Berci
Sep 10 '18 at 21:38
|
show 3 more comments
$begingroup$
The product $Xtimes Y$ implicitly comes with two fixed projections $pi_X,pi_Y$, and these play an implicit role in the definitions.
Dually, the coproduct $X+Y$ comes equipped with inclusions $iota_X,iota_Y$.
First, observe that (writing composition to the right),
$$underset{X+Y,to, Xtimes Y,to, X}{pmatrix{a&b\c&d}pi_X} = pmatrix{a\c}$$
and similarly, composing it by $pi_Y$ yields the second column.
Consequently, as $1_{Xtimes Y}=alpha,pmatrix{1&0\0&1}$, we have
$$ pi_X = alpha,pmatrix{1&0\0&1}pi_X =underset{Xtimes Yto X+Yto X}{qquad alpha,pmatrix{1\0}},. $$
Dual statements hold for rows and the $iota$'s.
From here, we can evaluate the 3 required elements like:
$$iota_Ycdot,pmatrix{1&f\0&1}alphapmatrix{1&g\0&1},cdotpi_X =
pmatrix{0&1},alpha,pmatrix{1\0} = pmatrix{0&1}pi_X = 0,.$$
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
$endgroup$
The product $Xtimes Y$ implicitly comes with two fixed projections $pi_X,pi_Y$, and these play an implicit role in the definitions.
Dually, the coproduct $X+Y$ comes equipped with inclusions $iota_X,iota_Y$.
First, observe that (writing composition to the right),
$$underset{X+Y,to, Xtimes Y,to, X}{pmatrix{a&b\c&d}pi_X} = pmatrix{a\c}$$
and similarly, composing it by $pi_Y$ yields the second column.
Consequently, as $1_{Xtimes Y}=alpha,pmatrix{1&0\0&1}$, we have
$$ pi_X = alpha,pmatrix{1&0\0&1}pi_X =underset{Xtimes Yto X+Yto X}{qquad alpha,pmatrix{1\0}},. $$
Dual statements hold for rows and the $iota$'s.
From here, we can evaluate the 3 required elements like:
$$iota_Ycdot,pmatrix{1&f\0&1}alphapmatrix{1&g\0&1},cdotpi_X =
pmatrix{0&1},alpha,pmatrix{1\0} = pmatrix{0&1}pi_X = 0,.$$
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
edited Sep 9 '18 at 23:58
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
answered Sep 9 '18 at 23:50
BerciBerci
62k23776
62k23776
1
$begingroup$
I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
$endgroup$
– user97678
Sep 10 '18 at 7:36
1
$begingroup$
Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
$endgroup$
– Berci
Sep 10 '18 at 7:46
1
$begingroup$
I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
$endgroup$
– user97678
Sep 10 '18 at 8:20
1
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
$endgroup$
– user97678
Sep 10 '18 at 17:52
1
$begingroup$
Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
$endgroup$
– Berci
Sep 10 '18 at 21:38
|
show 3 more comments
1
$begingroup$
I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
$endgroup$
– user97678
Sep 10 '18 at 7:36
1
$begingroup$
Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
$endgroup$
– Berci
Sep 10 '18 at 7:46
1
$begingroup$
I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
$endgroup$
– user97678
Sep 10 '18 at 8:20
1
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
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– user97678
Sep 10 '18 at 17:52
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Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
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– Berci
Sep 10 '18 at 21:38
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I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
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– user97678
Sep 10 '18 at 7:36
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I can follow your reasoning, but I am curious if you can generalize this to show that $pmatrix{a&b}alpha pmatrix{x\y}$ is $ax+by$ or something of the sort? (Defining $h = f + g$ in the example above).
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– user97678
Sep 10 '18 at 7:36
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Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
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– Berci
Sep 10 '18 at 7:46
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Yes, once distributivity over composition is proved. Check that we have $(a, b) =(a, 0)+(0,b)$ according to the definition, and then $$pmatrix{a&b}pmatrix{c\d} =pmatrix{a&0}pmatrix{c\d} +pmatrix{0 & b}pmatrix{c\d} =ac+bd$$
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– Berci
Sep 10 '18 at 7:46
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I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
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– user97678
Sep 10 '18 at 8:20
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I suppose I'm being dull here but I don't see how either of these statements are true... starting with $(a,0)+(0,b)$, presumably I want to look at its projections to show that they coincide with $(a,b)$'s, but I can't go from the definition of addition to understand how you can calculate the projections. I hope it's okay that I'm taking this beyond the scope of the original question--I'm struggling to get an intuition for these things.
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– user97678
Sep 10 '18 at 8:20
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I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
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– user97678
Sep 10 '18 at 17:52
$begingroup$
I can compute $pmatrix{1&(0,b)\0&1} pi_Y = pmatrix{b \ pi_Y}$ but I don't know how to cancel out the $alpha$ in the multiplication.
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– user97678
Sep 10 '18 at 17:52
1
1
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Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
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– Berci
Sep 10 '18 at 21:38
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Hmm.. Maybe the easiest way would be, if we could already fix all the $pi_X,pi_Y$ and $iota_X,iota_Y$ so that $pmatrix{1_X&0\0&1_Y}$ - hence $alpha$ also - eventually becomes the identity. Fix, say, the projections arbitrarily. Then, as $Xtimes Ycong X+Y$, the object $Xtimes Y$ itself can serve as the coproduct. (Basically we put the $alpha$'s in the inclusions $iota_X,iota_Y$.)
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– Berci
Sep 10 '18 at 21:38
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show 3 more comments
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This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A times B overset{alpha_{AB}}{to} A +B $$ denotes the inverse of the "identity matrix"
$$begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} $$
then we have the following identities:
$$ alpha_{AB} circ langle 1_A, 0_{AB} rangle = j_{A+B}^1 ,, quad alpha_{AB} circ langle 0_{BA}, 1_B rangle = j_{A+B}^2 ,, quad begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} = pi_{A times B}^1 ,, quad begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} = pi_{A times B}^2 ,.$$
The proofs follow from the definition of $alpha_{AB}$, namely:
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^1 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = alpha_{AB} circ langle 1_A, 0_{AB} rangle $$
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^2 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = alpha_{AB} circ langle 0_{BA} , 1_B rangle $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^1 = left( pi_{A times B}^1 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^1 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^1 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^2 = left( pi_{A times B}^2 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^2 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^2 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} $$ $triangle$
With that under our belt, the result is fairly straightforward:
$$ begin{bmatrix} 1_A & f \ 0_{BA} & 1_B end{bmatrix} cdot begin{bmatrix} 1_A & g \ 0_{BA} & 1_B end{bmatrix} = begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} circ begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} =: begin{bmatrix} h_{AA} & h_{AB} \ h_{BA} & h_{BB} end{bmatrix} ,.$$
Therefore we have that:
$$ h_{AA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 1_A, f rangle = pi_{Atimes B}^1 circ langle 1_A, f rangle = 1_A $$
$$ h_{BA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle =begin{array}{c} pi_{Atimes B}^1 circ langle 0_{BA}, 1_B rangle \ begin{cases}1_A \ 0_{BA} end{cases} circ j_{A+B}^2 end{array} = 0_{BA} $$
$$ h_{BB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} g \ 1_{B} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle = begin{cases} g \ 1_{B} end{cases} circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} g \ 1_B end{cases} circ alpha_{AB} circ langle 1_A, f rangle ,. $$
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
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add a comment |
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This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A times B overset{alpha_{AB}}{to} A +B $$ denotes the inverse of the "identity matrix"
$$begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} $$
then we have the following identities:
$$ alpha_{AB} circ langle 1_A, 0_{AB} rangle = j_{A+B}^1 ,, quad alpha_{AB} circ langle 0_{BA}, 1_B rangle = j_{A+B}^2 ,, quad begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} = pi_{A times B}^1 ,, quad begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} = pi_{A times B}^2 ,.$$
The proofs follow from the definition of $alpha_{AB}$, namely:
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^1 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = alpha_{AB} circ langle 1_A, 0_{AB} rangle $$
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^2 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = alpha_{AB} circ langle 0_{BA} , 1_B rangle $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^1 = left( pi_{A times B}^1 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^1 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^1 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^2 = left( pi_{A times B}^2 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^2 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^2 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} $$ $triangle$
With that under our belt, the result is fairly straightforward:
$$ begin{bmatrix} 1_A & f \ 0_{BA} & 1_B end{bmatrix} cdot begin{bmatrix} 1_A & g \ 0_{BA} & 1_B end{bmatrix} = begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} circ begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} =: begin{bmatrix} h_{AA} & h_{AB} \ h_{BA} & h_{BB} end{bmatrix} ,.$$
Therefore we have that:
$$ h_{AA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 1_A, f rangle = pi_{Atimes B}^1 circ langle 1_A, f rangle = 1_A $$
$$ h_{BA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle =begin{array}{c} pi_{Atimes B}^1 circ langle 0_{BA}, 1_B rangle \ begin{cases}1_A \ 0_{BA} end{cases} circ j_{A+B}^2 end{array} = 0_{BA} $$
$$ h_{BB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} g \ 1_{B} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle = begin{cases} g \ 1_{B} end{cases} circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} g \ 1_B end{cases} circ alpha_{AB} circ langle 1_A, f rangle ,. $$
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
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add a comment |
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This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A times B overset{alpha_{AB}}{to} A +B $$ denotes the inverse of the "identity matrix"
$$begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} $$
then we have the following identities:
$$ alpha_{AB} circ langle 1_A, 0_{AB} rangle = j_{A+B}^1 ,, quad alpha_{AB} circ langle 0_{BA}, 1_B rangle = j_{A+B}^2 ,, quad begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} = pi_{A times B}^1 ,, quad begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} = pi_{A times B}^2 ,.$$
The proofs follow from the definition of $alpha_{AB}$, namely:
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^1 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = alpha_{AB} circ langle 1_A, 0_{AB} rangle $$
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^2 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = alpha_{AB} circ langle 0_{BA} , 1_B rangle $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^1 = left( pi_{A times B}^1 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^1 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^1 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^2 = left( pi_{A times B}^2 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^2 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^2 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} $$ $triangle$
With that under our belt, the result is fairly straightforward:
$$ begin{bmatrix} 1_A & f \ 0_{BA} & 1_B end{bmatrix} cdot begin{bmatrix} 1_A & g \ 0_{BA} & 1_B end{bmatrix} = begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} circ begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} =: begin{bmatrix} h_{AA} & h_{AB} \ h_{BA} & h_{BB} end{bmatrix} ,.$$
Therefore we have that:
$$ h_{AA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 1_A, f rangle = pi_{Atimes B}^1 circ langle 1_A, f rangle = 1_A $$
$$ h_{BA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle =begin{array}{c} pi_{Atimes B}^1 circ langle 0_{BA}, 1_B rangle \ begin{cases}1_A \ 0_{BA} end{cases} circ j_{A+B}^2 end{array} = 0_{BA} $$
$$ h_{BB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} g \ 1_{B} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle = begin{cases} g \ 1_{B} end{cases} circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} g \ 1_B end{cases} circ alpha_{AB} circ langle 1_A, f rangle ,. $$
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
$endgroup$
This answer seems to be exactly the same as Berci's except worse. But I'll write the idea using Lawvere's notation so that maybe it is more clear to someone using the book (like me) learning this for the first time.
If the map $$ A times B overset{alpha_{AB}}{to} A +B $$ denotes the inverse of the "identity matrix"
$$begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} $$
then we have the following identities:
$$ alpha_{AB} circ langle 1_A, 0_{AB} rangle = j_{A+B}^1 ,, quad alpha_{AB} circ langle 0_{BA}, 1_B rangle = j_{A+B}^2 ,, quad begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} = pi_{A times B}^1 ,, quad begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} = pi_{A times B}^2 ,.$$
The proofs follow from the definition of $alpha_{AB}$, namely:
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^1 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = alpha_{AB} circ langle 1_A, 0_{AB} rangle $$
$$ alpha_{AB} circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} = 1_{A+B} implies j_{A+B}^2 = alpha_{AB} circ left( begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = alpha_{AB} circ langle 0_{BA} , 1_B rangle $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^1 = left( pi_{A times B}^1 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^1 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^1 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} $$
$$ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} = 1_{A times B} implies pi_{A times B}^2 = left( pi_{A times B}^2 circ begin{cases} langle 1_A, 0_{AB}rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = left( begin{cases} pi_{A times B}^2 circ langle 1_A, 0_{AB}rangle \ pi_{A times B}^2 circ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} = begin{cases} 0_{AB} \ 1_B end{cases} circ alpha_{AB} $$ $triangle$
With that under our belt, the result is fairly straightforward:
$$ begin{bmatrix} 1_A & f \ 0_{BA} & 1_B end{bmatrix} cdot begin{bmatrix} 1_A & g \ 0_{BA} & 1_B end{bmatrix} = begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} circ alpha_{AB} circ begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} =: begin{bmatrix} h_{AA} & h_{AB} \ h_{BA} & h_{BB} end{bmatrix} ,.$$
Therefore we have that:
$$ h_{AA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 1_A, f rangle = pi_{Atimes B}^1 circ langle 1_A, f rangle = 1_A $$
$$ h_{BA} = left( pi_{A+B}^1 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} 1_A \ 0_{BA} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle =begin{array}{c} pi_{Atimes B}^1 circ langle 0_{BA}, 1_B rangle \ begin{cases}1_A \ 0_{BA} end{cases} circ j_{A+B}^2 end{array} = 0_{BA} $$
$$ h_{BB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^2 right) = begin{cases} g \ 1_{B} end{cases} circ alpha_{AB} circ langle 0_{BA}, 1_B rangle = begin{cases} g \ 1_{B} end{cases} circ j_{A+B}^2 = 1_B$$
We even get as a bonus an explicit form for $f+g$ (although to be honest I have not yet found this helpful for solving the subsequent Exercise 1, which is why I wound up on Math.SE in the first place):
$$f+g := h_{AB} = left( pi_{A+B}^2 circ begin{cases} langle 1_A , g rangle \ langle 0_{BA}, 1_B rangle end{cases} right) circ alpha_{AB} circ left( begin{cases} langle 1_A , f rangle \ langle 0_{BA}, 1_B rangle end{cases} circ j_{A+B}^1 right) = begin{cases} g \ 1_B end{cases} circ alpha_{AB} circ langle 1_A, f rangle ,. $$
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
edited Jan 11 at 23:00
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
answered Jan 11 at 17:57
Chill2MachtChill2Macht
11.6k91869
11.6k91869
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– José Carlos Santos
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