What constructions of “elementary” mathematics are actually functors?












7












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I'm not looking for the usual simple examples of functors like the fundamental group or forgetful functors, what I'm looking for is some interesting examples of constructions from "elementary" mathematics that are secretly functorial. Like the derivative: it's actually a functor, with the chain rule expressing the composition rule of functors, but that's never discussed in basic Calculus courses.










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$endgroup$








  • 1




    $begingroup$
    What's the scope of what you call elementary mathematics ?
    $endgroup$
    – Max
    Jan 9 at 19:11






  • 3




    $begingroup$
    Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
    $endgroup$
    – Somos
    Jan 9 at 19:25






  • 1




    $begingroup$
    Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
    $endgroup$
    – violeta
    Jan 9 at 19:32






  • 1




    $begingroup$
    Do you want that to be discussed in a basic Calculus course?
    $endgroup$
    – Randall
    Jan 9 at 19:32






  • 1




    $begingroup$
    Not really... And about the scope, I left it open for interpretation intentionally.
    $endgroup$
    – violeta
    Jan 9 at 19:35
















7












$begingroup$


I'm not looking for the usual simple examples of functors like the fundamental group or forgetful functors, what I'm looking for is some interesting examples of constructions from "elementary" mathematics that are secretly functorial. Like the derivative: it's actually a functor, with the chain rule expressing the composition rule of functors, but that's never discussed in basic Calculus courses.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's the scope of what you call elementary mathematics ?
    $endgroup$
    – Max
    Jan 9 at 19:11






  • 3




    $begingroup$
    Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
    $endgroup$
    – Somos
    Jan 9 at 19:25






  • 1




    $begingroup$
    Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
    $endgroup$
    – violeta
    Jan 9 at 19:32






  • 1




    $begingroup$
    Do you want that to be discussed in a basic Calculus course?
    $endgroup$
    – Randall
    Jan 9 at 19:32






  • 1




    $begingroup$
    Not really... And about the scope, I left it open for interpretation intentionally.
    $endgroup$
    – violeta
    Jan 9 at 19:35














7












7








7


3



$begingroup$


I'm not looking for the usual simple examples of functors like the fundamental group or forgetful functors, what I'm looking for is some interesting examples of constructions from "elementary" mathematics that are secretly functorial. Like the derivative: it's actually a functor, with the chain rule expressing the composition rule of functors, but that's never discussed in basic Calculus courses.










share|cite|improve this question











$endgroup$




I'm not looking for the usual simple examples of functors like the fundamental group or forgetful functors, what I'm looking for is some interesting examples of constructions from "elementary" mathematics that are secretly functorial. Like the derivative: it's actually a functor, with the chain rule expressing the composition rule of functors, but that's never discussed in basic Calculus courses.







category-theory examples-counterexamples functors






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share|cite|improve this question













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share|cite|improve this question








edited Jan 9 at 19:03









Scientifica

6,80941335




6,80941335










asked Jan 9 at 18:59









violetavioleta

375210




375210








  • 1




    $begingroup$
    What's the scope of what you call elementary mathematics ?
    $endgroup$
    – Max
    Jan 9 at 19:11






  • 3




    $begingroup$
    Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
    $endgroup$
    – Somos
    Jan 9 at 19:25






  • 1




    $begingroup$
    Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
    $endgroup$
    – violeta
    Jan 9 at 19:32






  • 1




    $begingroup$
    Do you want that to be discussed in a basic Calculus course?
    $endgroup$
    – Randall
    Jan 9 at 19:32






  • 1




    $begingroup$
    Not really... And about the scope, I left it open for interpretation intentionally.
    $endgroup$
    – violeta
    Jan 9 at 19:35














  • 1




    $begingroup$
    What's the scope of what you call elementary mathematics ?
    $endgroup$
    – Max
    Jan 9 at 19:11






  • 3




    $begingroup$
    Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
    $endgroup$
    – Somos
    Jan 9 at 19:25






  • 1




    $begingroup$
    Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
    $endgroup$
    – violeta
    Jan 9 at 19:32






  • 1




    $begingroup$
    Do you want that to be discussed in a basic Calculus course?
    $endgroup$
    – Randall
    Jan 9 at 19:32






  • 1




    $begingroup$
    Not really... And about the scope, I left it open for interpretation intentionally.
    $endgroup$
    – violeta
    Jan 9 at 19:35








1




1




$begingroup$
What's the scope of what you call elementary mathematics ?
$endgroup$
– Max
Jan 9 at 19:11




$begingroup$
What's the scope of what you call elementary mathematics ?
$endgroup$
– Max
Jan 9 at 19:11




3




3




$begingroup$
Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
$endgroup$
– Somos
Jan 9 at 19:25




$begingroup$
Acutally, the derivative in elementary calculus is not a functor. If you consider a much more abstract setting as in Category of smooth manifolds then it is. Your example here is misleading.
$endgroup$
– Somos
Jan 9 at 19:25




1




1




$begingroup$
Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
$endgroup$
– violeta
Jan 9 at 19:32




$begingroup$
Sure, that's what I meant. To interpret anything as a functor we're gonna need to define the categories we're working on and do some abstract non-elementary work, that's okay.
$endgroup$
– violeta
Jan 9 at 19:32




1




1




$begingroup$
Do you want that to be discussed in a basic Calculus course?
$endgroup$
– Randall
Jan 9 at 19:32




$begingroup$
Do you want that to be discussed in a basic Calculus course?
$endgroup$
– Randall
Jan 9 at 19:32




1




1




$begingroup$
Not really... And about the scope, I left it open for interpretation intentionally.
$endgroup$
– violeta
Jan 9 at 19:35




$begingroup$
Not really... And about the scope, I left it open for interpretation intentionally.
$endgroup$
– violeta
Jan 9 at 19:35










5 Answers
5






active

oldest

votes


















11












$begingroup$

One of my favourite examples of these is group actions.



A (monoid or) group $G$ can be considered as a category with a single object $star$, whose morphisms $star to star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.



A left action of $G$ on a set $X$ is precisely a functor $alpha : G to mathbf{Set}$, where $G$ is considered as a category in the above sense.




  • The set $X$ is the value of $alpha(star)$;

  • For each $g in G$, we obtain a function $alpha_g = alpha(g) : X to X$;

  • Functoriality expresses the fact that $alpha_e = mathrm{id}_X$ and $alpha_{gh} = alpha_g circ alpha_h$.


Likewise a right action on a set is precisely a functor $alpha : G^{mathrm{op}} to mathbf{Set}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:31



















6












$begingroup$

The function that sends a set $X$ to its powerset $mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!



Probably the most natural way to make it into a functor is to define, given a function $f:Xto Y$,
$$mathcal{P}(f):mathcal{P}(X)to mathcal{P}(Y):Amapsto f(A)={f(a)mid ain A};$$
in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(gcirc f)(A)$ for all $f:Xto Y$, $g:Yto Z$ and $Asubset X$.



But there is also the function "inverse image by $f$", defined as
$$mathcal{P}'(f):mathcal{P}(Y)to mathcal{P}(X):Bmapsto f^{-1}(B)={ain Xmid f(a)in B}.$$
Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(gcirc f)^{-1}(C)$ for all $f:Xto Y$, $g:Yto Z$ and $Csubset Z$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
    $endgroup$
    – Jacob Maibach
    Jan 10 at 22:17



















2












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In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")






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$endgroup$









  • 1




    $begingroup$
    I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:35



















1












$begingroup$

Any homomorphism $f:G_1rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.



This is true for ring homomorphisms as well.






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$endgroup$









  • 1




    $begingroup$
    An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 22:58










  • $begingroup$
    Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 23:32



















0












$begingroup$

As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X times mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A times X$ described by @Randall (well, technically $X mapsto X times A$).



However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X to Y$, the derivative $D(f): TX to TY$ is the map
$$ D(f): (x,v) mapsto (f(x), df_{x}(v)) $$
where $df_{x}$ is the ordinary total derivative of $f$ at $x$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
    $endgroup$
    – Max
    Jan 10 at 23:00










  • $begingroup$
    @Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
    $endgroup$
    – Jacob Maibach
    Jan 10 at 23:49








  • 2




    $begingroup$
    Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
    $endgroup$
    – Max
    Jan 11 at 10:27










  • $begingroup$
    Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
    $endgroup$
    – Max
    Jan 11 at 10:28












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5 Answers
5






active

oldest

votes








5 Answers
5






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

One of my favourite examples of these is group actions.



A (monoid or) group $G$ can be considered as a category with a single object $star$, whose morphisms $star to star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.



A left action of $G$ on a set $X$ is precisely a functor $alpha : G to mathbf{Set}$, where $G$ is considered as a category in the above sense.




  • The set $X$ is the value of $alpha(star)$;

  • For each $g in G$, we obtain a function $alpha_g = alpha(g) : X to X$;

  • Functoriality expresses the fact that $alpha_e = mathrm{id}_X$ and $alpha_{gh} = alpha_g circ alpha_h$.


Likewise a right action on a set is precisely a functor $alpha : G^{mathrm{op}} to mathbf{Set}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:31
















11












$begingroup$

One of my favourite examples of these is group actions.



A (monoid or) group $G$ can be considered as a category with a single object $star$, whose morphisms $star to star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.



A left action of $G$ on a set $X$ is precisely a functor $alpha : G to mathbf{Set}$, where $G$ is considered as a category in the above sense.




  • The set $X$ is the value of $alpha(star)$;

  • For each $g in G$, we obtain a function $alpha_g = alpha(g) : X to X$;

  • Functoriality expresses the fact that $alpha_e = mathrm{id}_X$ and $alpha_{gh} = alpha_g circ alpha_h$.


Likewise a right action on a set is precisely a functor $alpha : G^{mathrm{op}} to mathbf{Set}$.






share|cite|improve this answer









$endgroup$









  • 4




    $begingroup$
    And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:31














11












11








11





$begingroup$

One of my favourite examples of these is group actions.



A (monoid or) group $G$ can be considered as a category with a single object $star$, whose morphisms $star to star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.



A left action of $G$ on a set $X$ is precisely a functor $alpha : G to mathbf{Set}$, where $G$ is considered as a category in the above sense.




  • The set $X$ is the value of $alpha(star)$;

  • For each $g in G$, we obtain a function $alpha_g = alpha(g) : X to X$;

  • Functoriality expresses the fact that $alpha_e = mathrm{id}_X$ and $alpha_{gh} = alpha_g circ alpha_h$.


Likewise a right action on a set is precisely a functor $alpha : G^{mathrm{op}} to mathbf{Set}$.






share|cite|improve this answer









$endgroup$



One of my favourite examples of these is group actions.



A (monoid or) group $G$ can be considered as a category with a single object $star$, whose morphisms $star to star$ are the elements of $G$, and whose identity and composition are given by the unit element $e$ and the group operation, respectively.



A left action of $G$ on a set $X$ is precisely a functor $alpha : G to mathbf{Set}$, where $G$ is considered as a category in the above sense.




  • The set $X$ is the value of $alpha(star)$;

  • For each $g in G$, we obtain a function $alpha_g = alpha(g) : X to X$;

  • Functoriality expresses the fact that $alpha_e = mathrm{id}_X$ and $alpha_{gh} = alpha_g circ alpha_h$.


Likewise a right action on a set is precisely a functor $alpha : G^{mathrm{op}} to mathbf{Set}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 19:13









Clive NewsteadClive Newstead

52k474136




52k474136








  • 4




    $begingroup$
    And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:31














  • 4




    $begingroup$
    And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:31








4




4




$begingroup$
And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
$endgroup$
– Marc Paul
Jan 10 at 22:31




$begingroup$
And if you replace Set with the category Vect of vector spaces over some field, you get representations of $G$.
$endgroup$
– Marc Paul
Jan 10 at 22:31











6












$begingroup$

The function that sends a set $X$ to its powerset $mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!



Probably the most natural way to make it into a functor is to define, given a function $f:Xto Y$,
$$mathcal{P}(f):mathcal{P}(X)to mathcal{P}(Y):Amapsto f(A)={f(a)mid ain A};$$
in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(gcirc f)(A)$ for all $f:Xto Y$, $g:Yto Z$ and $Asubset X$.



But there is also the function "inverse image by $f$", defined as
$$mathcal{P}'(f):mathcal{P}(Y)to mathcal{P}(X):Bmapsto f^{-1}(B)={ain Xmid f(a)in B}.$$
Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(gcirc f)^{-1}(C)$ for all $f:Xto Y$, $g:Yto Z$ and $Csubset Z$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
    $endgroup$
    – Jacob Maibach
    Jan 10 at 22:17
















6












$begingroup$

The function that sends a set $X$ to its powerset $mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!



Probably the most natural way to make it into a functor is to define, given a function $f:Xto Y$,
$$mathcal{P}(f):mathcal{P}(X)to mathcal{P}(Y):Amapsto f(A)={f(a)mid ain A};$$
in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(gcirc f)(A)$ for all $f:Xto Y$, $g:Yto Z$ and $Asubset X$.



But there is also the function "inverse image by $f$", defined as
$$mathcal{P}'(f):mathcal{P}(Y)to mathcal{P}(X):Bmapsto f^{-1}(B)={ain Xmid f(a)in B}.$$
Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(gcirc f)^{-1}(C)$ for all $f:Xto Y$, $g:Yto Z$ and $Csubset Z$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
    $endgroup$
    – Jacob Maibach
    Jan 10 at 22:17














6












6








6





$begingroup$

The function that sends a set $X$ to its powerset $mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!



Probably the most natural way to make it into a functor is to define, given a function $f:Xto Y$,
$$mathcal{P}(f):mathcal{P}(X)to mathcal{P}(Y):Amapsto f(A)={f(a)mid ain A};$$
in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(gcirc f)(A)$ for all $f:Xto Y$, $g:Yto Z$ and $Asubset X$.



But there is also the function "inverse image by $f$", defined as
$$mathcal{P}'(f):mathcal{P}(Y)to mathcal{P}(X):Bmapsto f^{-1}(B)={ain Xmid f(a)in B}.$$
Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(gcirc f)^{-1}(C)$ for all $f:Xto Y$, $g:Yto Z$ and $Csubset Z$.






share|cite|improve this answer









$endgroup$



The function that sends a set $X$ to its powerset $mathcal{P}(X)$ is a functor; and interestingly, it is a functor in more than one way!



Probably the most natural way to make it into a functor is to define, given a function $f:Xto Y$,
$$mathcal{P}(f):mathcal{P}(X)to mathcal{P}(Y):Amapsto f(A)={f(a)mid ain A};$$
in other words, we send $f$ to the function "direct image by $f$". This is a functor, because (pretty much by definition) $g(f(A))=(gcirc f)(A)$ for all $f:Xto Y$, $g:Yto Z$ and $Asubset X$.



But there is also the function "inverse image by $f$", defined as
$$mathcal{P}'(f):mathcal{P}(Y)to mathcal{P}(X):Bmapsto f^{-1}(B)={ain Xmid f(a)in B}.$$
Note that here I've switched $X$ and $Y$; so it's not a functor on the category of sets, but from the opposite category of sets to the category of sets, or if you prefer a contravariant functor on the category of sets. Here the functoriality amounts to the fact that $f^{-1}(g^{-1}(C))=(gcirc f)^{-1}(C)$ for all $f:Xto Y$, $g:Yto Z$ and $Csubset Z$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 22:02









Arnaud D.Arnaud D.

16.2k52445




16.2k52445








  • 1




    $begingroup$
    This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
    $endgroup$
    – Jacob Maibach
    Jan 10 at 22:17














  • 1




    $begingroup$
    This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
    $endgroup$
    – Jacob Maibach
    Jan 10 at 22:17








1




1




$begingroup$
This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
$endgroup$
– Jacob Maibach
Jan 10 at 22:17




$begingroup$
This is a great example since many basic properties of functions are naturally described in terms of the "direct image" map, such as how a function is injective if and only if its direct image map is cardinality-preserving.
$endgroup$
– Jacob Maibach
Jan 10 at 22:17











2












$begingroup$

In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:35
















2












$begingroup$

In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:35














2












2








2





$begingroup$

In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")






share|cite|improve this answer









$endgroup$



In almost any course a (U.S.) undergraduate math major would take, the construction $F(X)=A times X$ for fixed $A$ defines a functor from any reasonable category to itself (obvious action on maps). This works for sets, (abelian) groups, topological spaces, vector spaces, rings, etc. (Of course it works in any category with products, but I'm trying to keep it "elementary.")







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 19:40









RandallRandall

10.6k11431




10.6k11431








  • 1




    $begingroup$
    I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:35














  • 1




    $begingroup$
    I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
    $endgroup$
    – Marc Paul
    Jan 10 at 22:35








1




1




$begingroup$
I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
$endgroup$
– Marc Paul
Jan 10 at 22:35




$begingroup$
I think we don't even need to fix $A$, we can just view $times$ as a functor $mathcal{C} times mathcal{C} to mathcal{C}$.
$endgroup$
– Marc Paul
Jan 10 at 22:35











1












$begingroup$

Any homomorphism $f:G_1rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.



This is true for ring homomorphisms as well.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 22:58










  • $begingroup$
    Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 23:32
















1












$begingroup$

Any homomorphism $f:G_1rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.



This is true for ring homomorphisms as well.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 22:58










  • $begingroup$
    Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 23:32














1












1








1





$begingroup$

Any homomorphism $f:G_1rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.



This is true for ring homomorphisms as well.






share|cite|improve this answer









$endgroup$



Any homomorphism $f:G_1rightarrow G_2$ between two groups, both considered as one object categories, is a functor from $G_1$ to $G_2$.



This is true for ring homomorphisms as well.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 at 22:50









Oleg SmirnovOleg Smirnov

658




658








  • 1




    $begingroup$
    An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 22:58










  • $begingroup$
    Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 23:32














  • 1




    $begingroup$
    An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 22:58










  • $begingroup$
    Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
    $endgroup$
    – Oleg Smirnov
    Jan 18 at 23:32








1




1




$begingroup$
An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
$endgroup$
– Oleg Smirnov
Jan 18 at 22:58




$begingroup$
An interesting exercise is this: if two groups are equivalent as categories, are they isomorphic as groups?
$endgroup$
– Oleg Smirnov
Jan 18 at 22:58












$begingroup$
Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
$endgroup$
– Oleg Smirnov
Jan 18 at 23:32




$begingroup$
Perhaps a more "elementary" version of my suggestion is this: the procedure of evaluation of a polynomial $f(x)in{mathbb R}[x]$ at a real number $a$ is a functor from ${mathbb R}[x]$ to ${mathbb R}$, both considered as one object pre-additive categories.
$endgroup$
– Oleg Smirnov
Jan 18 at 23:32











0












$begingroup$

As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X times mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A times X$ described by @Randall (well, technically $X mapsto X times A$).



However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X to Y$, the derivative $D(f): TX to TY$ is the map
$$ D(f): (x,v) mapsto (f(x), df_{x}(v)) $$
where $df_{x}$ is the ordinary total derivative of $f$ at $x$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
    $endgroup$
    – Max
    Jan 10 at 23:00










  • $begingroup$
    @Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
    $endgroup$
    – Jacob Maibach
    Jan 10 at 23:49








  • 2




    $begingroup$
    Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
    $endgroup$
    – Max
    Jan 11 at 10:27










  • $begingroup$
    Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
    $endgroup$
    – Max
    Jan 11 at 10:28
















0












$begingroup$

As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X times mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A times X$ described by @Randall (well, technically $X mapsto X times A$).



However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X to Y$, the derivative $D(f): TX to TY$ is the map
$$ D(f): (x,v) mapsto (f(x), df_{x}(v)) $$
where $df_{x}$ is the ordinary total derivative of $f$ at $x$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
    $endgroup$
    – Max
    Jan 10 at 23:00










  • $begingroup$
    @Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
    $endgroup$
    – Jacob Maibach
    Jan 10 at 23:49








  • 2




    $begingroup$
    Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
    $endgroup$
    – Max
    Jan 11 at 10:27










  • $begingroup$
    Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
    $endgroup$
    – Max
    Jan 11 at 10:28














0












0








0





$begingroup$

As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X times mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A times X$ described by @Randall (well, technically $X mapsto X times A$).



However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X to Y$, the derivative $D(f): TX to TY$ is the map
$$ D(f): (x,v) mapsto (f(x), df_{x}(v)) $$
where $df_{x}$ is the ordinary total derivative of $f$ at $x$.






share|cite|improve this answer











$endgroup$



As noted, differentiation is a functor in the category of (real, finite-dimensional) smooth manifolds. In particular, it maps every manifold $X$ to its tangent bundle $TX$, which is locally isomorphic to $X times mathbb{R}^{n}$ for appropriate $n$. In some sense, one might consider differentiation ''locally'' an example functor of the form $F(X) = A times X$ described by @Randall (well, technically $X mapsto X times A$).



However, the functor maps smooth functions in a non-trivial way. For manifolds $X, Y$ and smooth $f: X to Y$, the derivative $D(f): TX to TY$ is the map
$$ D(f): (x,v) mapsto (f(x), df_{x}(v)) $$
where $df_{x}$ is the ordinary total derivative of $f$ at $x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 5 at 5:08

























answered Jan 10 at 21:43









Jacob MaibachJacob Maibach

1,4902917




1,4902917








  • 1




    $begingroup$
    $TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
    $endgroup$
    – Max
    Jan 10 at 23:00










  • $begingroup$
    @Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
    $endgroup$
    – Jacob Maibach
    Jan 10 at 23:49








  • 2




    $begingroup$
    Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
    $endgroup$
    – Max
    Jan 11 at 10:27










  • $begingroup$
    Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
    $endgroup$
    – Max
    Jan 11 at 10:28














  • 1




    $begingroup$
    $TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
    $endgroup$
    – Max
    Jan 10 at 23:00










  • $begingroup$
    @Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
    $endgroup$
    – Jacob Maibach
    Jan 10 at 23:49








  • 2




    $begingroup$
    Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
    $endgroup$
    – Max
    Jan 11 at 10:27










  • $begingroup$
    Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
    $endgroup$
    – Max
    Jan 11 at 10:28








1




1




$begingroup$
$TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
$endgroup$
– Max
Jan 10 at 23:00




$begingroup$
$TX$ is not isomorphic to $Xtimes mathbb{R}^n$ in general.
$endgroup$
– Max
Jan 10 at 23:00












$begingroup$
@Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
$endgroup$
– Jacob Maibach
Jan 10 at 23:49






$begingroup$
@Max How so? Is it the topology? I thought at least locally it is isomorphic (which I admit is quite different than being globally isomorphic).
$endgroup$
– Jacob Maibach
Jan 10 at 23:49






2




2




$begingroup$
Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
$endgroup$
– Max
Jan 11 at 10:27




$begingroup$
Yes it is locally isomorphic but not globally; but imagine if it were, then vector fields would be much less interesting : a vector field would be essentially a map $Xto mathbb{R}^n$, and that would be boring. Manifolds for which this isomorphism holds are called parallelisable (at least in french, I don't know the english word). It's equivalent to having $n$ vector fields $X_1,...,X_n$ such that for all $x$, $(X_1(x),...,X_n(x))$ is a basis of $T_xX$. In particulaf, a manifold for which there is such an isomorphism are always orientable (but it's not equivalent) : $mathbb{R}P^2$ isn't
$endgroup$
– Max
Jan 11 at 10:27












$begingroup$
Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
$endgroup$
– Max
Jan 11 at 10:28




$begingroup$
Butnit doesn't change the fact that $Xmapsto TX$ is a functor, simply i's a more interesting one
$endgroup$
– Max
Jan 11 at 10:28


















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