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Show that the disjoint union is the coproduct in the category of topological spaces
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Given any family $X_alpha_alpha in A$ of topological spaces, show that the disjoint union space $bigsqcup_alpha in A X_alpha$ is their coproduct in the category Top
My attemtped proof: Fix $alpha in A$. The disjoint union space comes together with a continuous inclusion map $i_alpha : X_alpha to bigsqcup_alpha in AX_alpha$. Let $W$ be any topological space and let $f_alpha : X_alpha to W$ be any continuous map.
We need to show that there exists a unique continuous map $f: bigsqcup_alpha in AX_alpha to W$ such that $f circ i_alpha = f_alpha$. To that end define $f : bigsqcup_alpha in AX_alpha to W$ by $$f(x, alpha) = f_alpha(x).$$By the characteristic property of the disjoint union space, $f$ is continuous if and only if its restriction to each $(X_alpha)^* = i_alpha[X_alpha]$ is continuous. So it suffices to show that such a restriction is continuous to show continuity of $f$.
With that in mind consider the restriction $f|_(X_alpha)^* : (X_alpha)^* to W$ and define $pi_alpha : i_alpha[X_alpha] to X_alpha$ by $pi(x, alpha) = x$. Note that $pi_alpha$ is continuous and we have $$f|_(X_alpha)^* = f_alpha circ pi_alpha$$ and since $f_alpha $ and $pi_alpha$ are continuous we have that $f|_(X_alpha)^*$ is continuous. Since $alpha$ was fixed arbitrarily this holds for all $alpha in A$. Hence by the characteristic property we conclude that $f$ is continuous.
Now I have to show that the map $f$ is unique. However I'm not sure how to proceed with that. How can I prove that $f$ is unique? Also is what I've written above correct?
general-topology category-theory
asked Jul 28 at 15:09
Perturbative
3,47911039
add a comment |Â
up vote
1
down vote
favorite
Given any family $X_alpha_alpha in A$ of topological spaces, show that the disjoint union space $bigsqcup_alpha in A X_alpha$ is their coproduct in the category Top
My attemtped proof: Fix $alpha in A$. The disjoint union space comes together with a continuous inclusion map $i_alpha : X_alpha to bigsqcup_alpha in AX_alpha$. Let $W$ be any topological space and let $f_alpha : X_alpha to W$ be any continuous map.
We need to show that there exists a unique continuous map $f: bigsqcup_alpha in AX_alpha to W$ such that $f circ i_alpha = f_alpha$. To that end define $f : bigsqcup_alpha in AX_alpha to W$ by $$f(x, alpha) = f_alpha(x).$$By the characteristic property of the disjoint union space, $f$ is continuous if and only if its restriction to each $(X_alpha)^* = i_alpha[X_alpha]$ is continuous. So it suffices to show that such a restriction is continuous to show continuity of $f$.
With that in mind consider the restriction $f|_(X_alpha)^* : (X_alpha)^* to W$ and define $pi_alpha : i_alpha[X_alpha] to X_alpha$ by $pi(x, alpha) = x$. Note that $pi_alpha$ is continuous and we have $$f|_(X_alpha)^* = f_alpha circ pi_alpha$$ and since $f_alpha $ and $pi_alpha$ are continuous we have that $f|_(X_alpha)^*$ is continuous. Since $alpha$ was fixed arbitrarily this holds for all $alpha in A$. Hence by the characteristic property we conclude that $f$ is continuous.
Now I have to show that the map $f$ is unique. However I'm not sure how to proceed with that. How can I prove that $f$ is unique? Also is what I've written above correct?
general-topology category-theory
asked Jul 28 at 15:09
Perturbative
3,47911039
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given any family $X_alpha_alpha in A$ of topological spaces, show that the disjoint union space $bigsqcup_alpha in A X_alpha$ is their coproduct in the category Top
My attemtped proof: Fix $alpha in A$. The disjoint union space comes together with a continuous inclusion map $i_alpha : X_alpha to bigsqcup_alpha in AX_alpha$. Let $W$ be any topological space and let $f_alpha : X_alpha to W$ be any continuous map.
We need to show that there exists a unique continuous map $f: bigsqcup_alpha in AX_alpha to W$ such that $f circ i_alpha = f_alpha$. To that end define $f : bigsqcup_alpha in AX_alpha to W$ by $$f(x, alpha) = f_alpha(x).$$By the characteristic property of the disjoint union space, $f$ is continuous if and only if its restriction to each $(X_alpha)^* = i_alpha[X_alpha]$ is continuous. So it suffices to show that such a restriction is continuous to show continuity of $f$.
With that in mind consider the restriction $f|_(X_alpha)^* : (X_alpha)^* to W$ and define $pi_alpha : i_alpha[X_alpha] to X_alpha$ by $pi(x, alpha) = x$. Note that $pi_alpha$ is continuous and we have $$f|_(X_alpha)^* = f_alpha circ pi_alpha$$ and since $f_alpha $ and $pi_alpha$ are continuous we have that $f|_(X_alpha)^*$ is continuous. Since $alpha$ was fixed arbitrarily this holds for all $alpha in A$. Hence by the characteristic property we conclude that $f$ is continuous.
Now I have to show that the map $f$ is unique. However I'm not sure how to proceed with that. How can I prove that $f$ is unique? Also is what I've written above correct?
general-topology category-theory
asked Jul 28 at 15:09
Perturbative
3,47911039
Given any family $X_alpha_alpha in A$ of topological spaces, show that the disjoint union space $bigsqcup_alpha in A X_alpha$ is their coproduct in the category Top
My attemtped proof: Fix $alpha in A$. The disjoint union space comes together with a continuous inclusion map $i_alpha : X_alpha to bigsqcup_alpha in AX_alpha$. Let $W$ be any topological space and let $f_alpha : X_alpha to W$ be any continuous map.
We need to show that there exists a unique continuous map $f: bigsqcup_alpha in AX_alpha to W$ such that $f circ i_alpha = f_alpha$. To that end define $f : bigsqcup_alpha in AX_alpha to W$ by $$f(x, alpha) = f_alpha(x).$$By the characteristic property of the disjoint union space, $f$ is continuous if and only if its restriction to each $(X_alpha)^* = i_alpha[X_alpha]$ is continuous. So it suffices to show that such a restriction is continuous to show continuity of $f$.
With that in mind consider the restriction $f|_(X_alpha)^* : (X_alpha)^* to W$ and define $pi_alpha : i_alpha[X_alpha] to X_alpha$ by $pi(x, alpha) = x$. Note that $pi_alpha$ is continuous and we have $$f|_(X_alpha)^* = f_alpha circ pi_alpha$$ and since $f_alpha $ and $pi_alpha$ are continuous we have that $f|_(X_alpha)^*$ is continuous. Since $alpha$ was fixed arbitrarily this holds for all $alpha in A$. Hence by the characteristic property we conclude that $f$ is continuous.
Now I have to show that the map $f$ is unique. However I'm not sure how to proceed with that. How can I prove that $f$ is unique? Also is what I've written above correct?
general-topology category-theory
asked Jul 28 at 15:09
Perturbative
3,47911039
asked Jul 28 at 15:09
Perturbative
3,47911039
asked Jul 28 at 15:09
Perturbative
3,47911039
asked Jul 28 at 15:09
Perturbative
3,47911039
3,47911039
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3 Answers
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You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.
As for uniqueness, that's easy. The only function $f$ such that $f circ i_alpha = f_alpha$ has to be given on each element $(x, alpha)$ by $f(x, alpha)=f(i_alpha(x)) = f_alpha(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.
answered Jul 28 at 15:25
Chessanator
1,592210
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
add a comment |Â
up vote
1
down vote
A typical point of the coproduct is $i_alpha(x)in i_alpha[X_alpha]$ for some $alpha$ with $xin X_alpha$. Then we must have
$f(i_alpha(x))=f_alpha(x)$. Therefore there is only one possible
value for each $f(y)$ for $yincoprod_alpha X_alpha$.
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
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up vote
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This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : mathbfTop to mathbfSet$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_alpha)_alphain A$ is the disjoint union of the underlying sets $(U(X_alpha))_alphain A$. Now you just have to find the topology you need to put on $coprod_alpha U(X_alpha)$ in order to make it the disjoint union in $mathbfTop$. But the open sets of any topological space $X$ are identified with the continuous map $Xto mathbb S$, where $mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $0$ open and $1$ not open. In particular, whatever $coprod_alpha X_alpha$ is, if it exists, its set of open sets is in bijection with
$$mathbfTop(coprod_alpha X_alpha,mathbb S)simeq prod_alphamathbfTop(X_alpha,mathbb S)$$
The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $coprod_alpha X_alpha$ is exactly given by an open set in each of the $X_alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $iota_beta:X_beta to coprod_alpha X_alpha$. So the bijection is actually saying that $coprod_alpha X_alpha$ has the final topology given by the maps $U(iota_beta)$. Taking into account the remark we made about $U(coprod_alpha X_alpha)$, it is actually the definition of the disjoint union topology.
answered Jul 28 at 17:33
Pece
7,92211040
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.
As for uniqueness, that's easy. The only function $f$ such that $f circ i_alpha = f_alpha$ has to be given on each element $(x, alpha)$ by $f(x, alpha)=f(i_alpha(x)) = f_alpha(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.
answered Jul 28 at 15:25
Chessanator
1,592210
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
add a comment |Â
up vote
1
down vote
accepted
You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.
As for uniqueness, that's easy. The only function $f$ such that $f circ i_alpha = f_alpha$ has to be given on each element $(x, alpha)$ by $f(x, alpha)=f(i_alpha(x)) = f_alpha(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.
answered Jul 28 at 15:25
Chessanator
1,592210
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.
As for uniqueness, that's easy. The only function $f$ such that $f circ i_alpha = f_alpha$ has to be given on each element $(x, alpha)$ by $f(x, alpha)=f(i_alpha(x)) = f_alpha(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.
answered Jul 28 at 15:25
Chessanator
1,592210
You'd need to check with your lecturer, but I'm a bit dubious about your phrasing "By the characteristic property of the disjoint union space." However you go about stating that characteristic property, it's awfully close to just saying "The characteristic property of the disjoint union is that it's the coproduct." It depends on what other information you have available, but even then for myself I'd prefer to prove continuity of $f$ by checking the pre-image of each open set.
As for uniqueness, that's easy. The only function $f$ such that $f circ i_alpha = f_alpha$ has to be given on each element $(x, alpha)$ by $f(x, alpha)=f(i_alpha(x)) = f_alpha(x)$ so is unique. In hindsight, I'd put that at the start of your proof as the way of identifying $f$ as the function you had to pick.
answered Jul 28 at 15:25
Chessanator
1,592210
answered Jul 28 at 15:25
Chessanator
1,592210
answered Jul 28 at 15:25
Chessanator
1,592210
answered Jul 28 at 15:25
Chessanator
1,592210
1,592210
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
add a comment |Â
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
I'm working out of Introduction to Topological Manifolds by John Lee and in his book the characteristic property of the disjoint union space is given before this problem so I'm assuming that there isn't any circular reasoning here.
– Perturbative
Jul 28 at 15:53
1
1
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
That's reasonable. In that case, I'd just familiarise myself with his proof of the characteristic property and recognise that's where most of the mathematical work is. (Hell, for all I know, Lee's purpose in including this question at this point and in this way was to trigger the realisation 'The specific definition for topological disjoint union is the same as the category-theory notion of coproduct'. In which case, you've engaged with the problem in an ideal way by quickly realising that connection.
– Chessanator
Jul 28 at 16:04
add a comment |Â
up vote
1
down vote
A typical point of the coproduct is $i_alpha(x)in i_alpha[X_alpha]$ for some $alpha$ with $xin X_alpha$. Then we must have
$f(i_alpha(x))=f_alpha(x)$. Therefore there is only one possible
value for each $f(y)$ for $yincoprod_alpha X_alpha$.
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
1
down vote
A typical point of the coproduct is $i_alpha(x)in i_alpha[X_alpha]$ for some $alpha$ with $xin X_alpha$. Then we must have
$f(i_alpha(x))=f_alpha(x)$. Therefore there is only one possible
value for each $f(y)$ for $yincoprod_alpha X_alpha$.
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A typical point of the coproduct is $i_alpha(x)in i_alpha[X_alpha]$ for some $alpha$ with $xin X_alpha$. Then we must have
$f(i_alpha(x))=f_alpha(x)$. Therefore there is only one possible
value for each $f(y)$ for $yincoprod_alpha X_alpha$.
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
A typical point of the coproduct is $i_alpha(x)in i_alpha[X_alpha]$ for some $alpha$ with $xin X_alpha$. Then we must have
$f(i_alpha(x))=f_alpha(x)$. Therefore there is only one possible
value for each $f(y)$ for $yincoprod_alpha X_alpha$.
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 15:19
Lord Shark the Unknown
84.6k950111
84.6k950111
add a comment |Â
add a comment |Â
up vote
1
down vote
This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : mathbfTop to mathbfSet$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_alpha)_alphain A$ is the disjoint union of the underlying sets $(U(X_alpha))_alphain A$. Now you just have to find the topology you need to put on $coprod_alpha U(X_alpha)$ in order to make it the disjoint union in $mathbfTop$. But the open sets of any topological space $X$ are identified with the continuous map $Xto mathbb S$, where $mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $0$ open and $1$ not open. In particular, whatever $coprod_alpha X_alpha$ is, if it exists, its set of open sets is in bijection with
$$mathbfTop(coprod_alpha X_alpha,mathbb S)simeq prod_alphamathbfTop(X_alpha,mathbb S)$$
The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $coprod_alpha X_alpha$ is exactly given by an open set in each of the $X_alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $iota_beta:X_beta to coprod_alpha X_alpha$. So the bijection is actually saying that $coprod_alpha X_alpha$ has the final topology given by the maps $U(iota_beta)$. Taking into account the remark we made about $U(coprod_alpha X_alpha)$, it is actually the definition of the disjoint union topology.
answered Jul 28 at 17:33
Pece
7,92211040
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
add a comment |Â
up vote
1
down vote
This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : mathbfTop to mathbfSet$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_alpha)_alphain A$ is the disjoint union of the underlying sets $(U(X_alpha))_alphain A$. Now you just have to find the topology you need to put on $coprod_alpha U(X_alpha)$ in order to make it the disjoint union in $mathbfTop$. But the open sets of any topological space $X$ are identified with the continuous map $Xto mathbb S$, where $mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $0$ open and $1$ not open. In particular, whatever $coprod_alpha X_alpha$ is, if it exists, its set of open sets is in bijection with
$$mathbfTop(coprod_alpha X_alpha,mathbb S)simeq prod_alphamathbfTop(X_alpha,mathbb S)$$
The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $coprod_alpha X_alpha$ is exactly given by an open set in each of the $X_alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $iota_beta:X_beta to coprod_alpha X_alpha$. So the bijection is actually saying that $coprod_alpha X_alpha$ has the final topology given by the maps $U(iota_beta)$. Taking into account the remark we made about $U(coprod_alpha X_alpha)$, it is actually the definition of the disjoint union topology.
answered Jul 28 at 17:33
Pece
7,92211040
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : mathbfTop to mathbfSet$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_alpha)_alphain A$ is the disjoint union of the underlying sets $(U(X_alpha))_alphain A$. Now you just have to find the topology you need to put on $coprod_alpha U(X_alpha)$ in order to make it the disjoint union in $mathbfTop$. But the open sets of any topological space $X$ are identified with the continuous map $Xto mathbb S$, where $mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $0$ open and $1$ not open. In particular, whatever $coprod_alpha X_alpha$ is, if it exists, its set of open sets is in bijection with
$$mathbfTop(coprod_alpha X_alpha,mathbb S)simeq prod_alphamathbfTop(X_alpha,mathbb S)$$
The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $coprod_alpha X_alpha$ is exactly given by an open set in each of the $X_alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $iota_beta:X_beta to coprod_alpha X_alpha$. So the bijection is actually saying that $coprod_alpha X_alpha$ has the final topology given by the maps $U(iota_beta)$. Taking into account the remark we made about $U(coprod_alpha X_alpha)$, it is actually the definition of the disjoint union topology.
answered Jul 28 at 17:33
Pece
7,92211040
This is a very good exercise to do it by hand as you attempted to and others have already provided some help. I just want to complete these by proposing a very "abstract nonsense" proof.
The forgetful functor $U : mathbfTop to mathbfSet$ admits a right adjoint, namely the functor that sends a set $X$ to the space $X$ with trivial topology (some call it indiscrete topology also i think). So the forgetful functor $U$ commutes with colimits, meaning in particular that the underlying set of the disjoint union of topological spaces $(X_alpha)_alphain A$ is the disjoint union of the underlying sets $(U(X_alpha))_alphain A$. Now you just have to find the topology you need to put on $coprod_alpha U(X_alpha)$ in order to make it the disjoint union in $mathbfTop$. But the open sets of any topological space $X$ are identified with the continuous map $Xto mathbb S$, where $mathbb S$ is the Sierpinski space : it has two elements $0$ and $1$ with $0$ open and $1$ not open. In particular, whatever $coprod_alpha X_alpha$ is, if it exists, its set of open sets is in bijection with
$$mathbfTop(coprod_alpha X_alpha,mathbb S)simeq prod_alphamathbfTop(X_alpha,mathbb S)$$
The right hand side is just a use of the universal property of the coproduct. Hence it says that an open set of $coprod_alpha X_alpha$ is exactly given by an open set in each of the $X_alpha$'s. Now you have to be a little careful: the bijection just above is given by precomposition with the canonical maps $iota_beta:X_beta to coprod_alpha X_alpha$. So the bijection is actually saying that $coprod_alpha X_alpha$ has the final topology given by the maps $U(iota_beta)$. Taking into account the remark we made about $U(coprod_alpha X_alpha)$, it is actually the definition of the disjoint union topology.
answered Jul 28 at 17:33
Pece
7,92211040
answered Jul 28 at 17:33
Pece
7,92211040
answered Jul 28 at 17:33
Pece
7,92211040
answered Jul 28 at 17:33
Pece
7,92211040
7,92211040
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
add a comment |Â
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
Thank you for this answer!
– Perturbative
Jul 28 at 21:44
add a comment |Â
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