Proof that direct product of connected spaces is connected
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I am trying to understand a piece of the proof that the direct product of connected spaces is itself a connected space, as given by Lee in "Introduction to Topological Manifolds". By induction, we may consider the case with two spaces.
So suppose that $X$ and $Y$ are connected topological spaces, and suppose for sake of contradiction that there exist open sets $U$ and $V$ that disconnect $Xtimes Y$. Let $(x_0,y_0)$ be a point in $Y$.
The next part is where I am confused. The author writes "The set ${x_0}times Y$ is connected because it is homeomorphic to $Y$. Why is This space homeomorphic to $Y$? I don't see how there can be an injective map from ${x_0}times Y$ to $Y$. What am I missing? Is there some obvious homeomorphism?
general-topology connectedness
asked Jan 10 at 19:59
confusedmath confusedmath
24919
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add a comment |
$begingroup$
I am trying to understand a piece of the proof that the direct product of connected spaces is itself a connected space, as given by Lee in "Introduction to Topological Manifolds". By induction, we may consider the case with two spaces.
So suppose that $X$ and $Y$ are connected topological spaces, and suppose for sake of contradiction that there exist open sets $U$ and $V$ that disconnect $Xtimes Y$. Let $(x_0,y_0)$ be a point in $Y$.
The next part is where I am confused. The author writes "The set ${x_0}times Y$ is connected because it is homeomorphic to $Y$. Why is This space homeomorphic to $Y$? I don't see how there can be an injective map from ${x_0}times Y$ to $Y$. What am I missing? Is there some obvious homeomorphism?
general-topology connectedness
asked Jan 10 at 19:59
confusedmath confusedmath
24919
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2
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The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
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– Will M.
Jan 10 at 20:00
2
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$f(x_0,y)=y$ is injective
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– Randall
Jan 10 at 20:00
2
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00
add a comment |
$begingroup$
I am trying to understand a piece of the proof that the direct product of connected spaces is itself a connected space, as given by Lee in "Introduction to Topological Manifolds". By induction, we may consider the case with two spaces.
So suppose that $X$ and $Y$ are connected topological spaces, and suppose for sake of contradiction that there exist open sets $U$ and $V$ that disconnect $Xtimes Y$. Let $(x_0,y_0)$ be a point in $Y$.
The next part is where I am confused. The author writes "The set ${x_0}times Y$ is connected because it is homeomorphic to $Y$. Why is This space homeomorphic to $Y$? I don't see how there can be an injective map from ${x_0}times Y$ to $Y$. What am I missing? Is there some obvious homeomorphism?
general-topology connectedness
asked Jan 10 at 19:59
confusedmath confusedmath
24919
$endgroup$
I am trying to understand a piece of the proof that the direct product of connected spaces is itself a connected space, as given by Lee in "Introduction to Topological Manifolds". By induction, we may consider the case with two spaces.
So suppose that $X$ and $Y$ are connected topological spaces, and suppose for sake of contradiction that there exist open sets $U$ and $V$ that disconnect $Xtimes Y$. Let $(x_0,y_0)$ be a point in $Y$.
The next part is where I am confused. The author writes "The set ${x_0}times Y$ is connected because it is homeomorphic to $Y$. Why is This space homeomorphic to $Y$? I don't see how there can be an injective map from ${x_0}times Y$ to $Y$. What am I missing? Is there some obvious homeomorphism?
general-topology connectedness
general-topology connectedness
asked Jan 10 at 19:59
confusedmath confusedmath
24919
asked Jan 10 at 19:59
confusedmath confusedmath
24919
asked Jan 10 at 19:59
confusedmath confusedmath
24919
asked Jan 10 at 19:59
confusedmath confusedmath
24919
asked Jan 10 at 19:59
confusedmath confusedmath
24919
24919
2
$begingroup$
The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
$endgroup$
– Will M.
Jan 10 at 20:00
2
$begingroup$
$f(x_0,y)=y$ is injective
$endgroup$
– Randall
Jan 10 at 20:00
2
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00
add a comment |
2
$begingroup$
The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
$endgroup$
– Will M.
Jan 10 at 20:00
2
$begingroup$
$f(x_0,y)=y$ is injective
$endgroup$
– Randall
Jan 10 at 20:00
2
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00
2
2
$begingroup$
The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
$endgroup$
– Will M.
Jan 10 at 20:00
$begingroup$
The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
$endgroup$
– Will M.
Jan 10 at 20:00
2
2
$begingroup$
$f(x_0,y)=y$ is injective
$endgroup$
– Randall
Jan 10 at 20:00
$begingroup$
$f(x_0,y)=y$ is injective
$endgroup$
– Randall
Jan 10 at 20:00
2
2
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00
add a comment |
1 Answer
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Take the map $phi : {x_0} times Y to Y$ that maps $(x_0, y) mapsto y$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $phi^{-1}$ is the map which takes $y mapsto(x_0, y)$ is also clearly continuous, which tells us that $phi$ is an open mapping. Thus we have a homeomorphism.
answered Jan 10 at 20:00
TrostAftTrostAft
433412
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add a comment |
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1 Answer
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$begingroup$
Take the map $phi : {x_0} times Y to Y$ that maps $(x_0, y) mapsto y$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $phi^{-1}$ is the map which takes $y mapsto(x_0, y)$ is also clearly continuous, which tells us that $phi$ is an open mapping. Thus we have a homeomorphism.
answered Jan 10 at 20:00
TrostAftTrostAft
433412
$endgroup$
add a comment |
$begingroup$
Take the map $phi : {x_0} times Y to Y$ that maps $(x_0, y) mapsto y$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $phi^{-1}$ is the map which takes $y mapsto(x_0, y)$ is also clearly continuous, which tells us that $phi$ is an open mapping. Thus we have a homeomorphism.
answered Jan 10 at 20:00
TrostAftTrostAft
433412
$endgroup$
add a comment |
$begingroup$
Take the map $phi : {x_0} times Y to Y$ that maps $(x_0, y) mapsto y$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $phi^{-1}$ is the map which takes $y mapsto(x_0, y)$ is also clearly continuous, which tells us that $phi$ is an open mapping. Thus we have a homeomorphism.
answered Jan 10 at 20:00
TrostAftTrostAft
433412
$endgroup$
Take the map $phi : {x_0} times Y to Y$ that maps $(x_0, y) mapsto y$. It's clear that this is a bijective continuous function. Furthermore, it's inverse $phi^{-1}$ is the map which takes $y mapsto(x_0, y)$ is also clearly continuous, which tells us that $phi$ is an open mapping. Thus we have a homeomorphism.
answered Jan 10 at 20:00
TrostAftTrostAft
433412
answered Jan 10 at 20:00
TrostAftTrostAft
433412
answered Jan 10 at 20:00
TrostAftTrostAft
433412
answered Jan 10 at 20:00
TrostAftTrostAft
433412
433412
add a comment |
add a comment |
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The obvious map from ${x_0} times mathrm{Y}$ to $mathrm{Y}$ is $(x_0, y) mapsto y,$ which, is inverse of $y mapsto (x_0, y).$
$endgroup$
– Will M.
Jan 10 at 20:00
2
$begingroup$
$f(x_0,y)=y$ is injective
$endgroup$
– Randall
Jan 10 at 20:00
2
$begingroup$
You're likely missing that $x_0$ is fixed.
$endgroup$
– Randall
Jan 10 at 20:00