Why is a categorical product in Top uniquely equal to the product topology?
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My question is NOT “why does the product topology satisfy the conditions of a categorical product in Top”.
Rather, my question is, why does it do so uniquely? I don’t require necessarily a fully rigorous proof, but an conceptual understanding.
My guess is: If we take the cartesian product between two topological spaces $X,Y$, but we endow this product with a trivial topology ${emptyset, Xtimes Y}$, then this also satisfies the categorical product conditions:
For any topological space $Z$ and continuous functions $f:Zto X, g:Zto Y$, there is a unique function $h=ftimes g:Zto Xtimes Y$. We don’t need to put any more conditions on the topology of $Xtimes Y$ for $h$ to be unique. (This is essentially because the cartesian product $Xtimes Y$ is already “set-theoretically restricted” enough to ensure uniqueness of $h$). Therefore, we can put any topology on $Xtimes Y$, so long as it makes $h$ unique for all such $(Z,f,g)$. One topology that does this is the product topology, but another one that does it is the topology ${emptyset,Xtimes Y }$.
What’s wrong with my argument?
EDIT: I Immediately realized that while this topology gets you a continuous $h$, it doesn’t get you continuous projections $pi_X,pi_Y$...
general-topology category-theory
asked Nov 21 at 12:21
user56834
3,13821149
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show 1 more comment
up vote
-1
down vote
favorite
My question is NOT “why does the product topology satisfy the conditions of a categorical product in Top”.
Rather, my question is, why does it do so uniquely? I don’t require necessarily a fully rigorous proof, but an conceptual understanding.
My guess is: If we take the cartesian product between two topological spaces $X,Y$, but we endow this product with a trivial topology ${emptyset, Xtimes Y}$, then this also satisfies the categorical product conditions:
For any topological space $Z$ and continuous functions $f:Zto X, g:Zto Y$, there is a unique function $h=ftimes g:Zto Xtimes Y$. We don’t need to put any more conditions on the topology of $Xtimes Y$ for $h$ to be unique. (This is essentially because the cartesian product $Xtimes Y$ is already “set-theoretically restricted” enough to ensure uniqueness of $h$). Therefore, we can put any topology on $Xtimes Y$, so long as it makes $h$ unique for all such $(Z,f,g)$. One topology that does this is the product topology, but another one that does it is the topology ${emptyset,Xtimes Y }$.
What’s wrong with my argument?
EDIT: I Immediately realized that while this topology gets you a continuous $h$, it doesn’t get you continuous projections $pi_X,pi_Y$...
general-topology category-theory
asked Nov 21 at 12:21
user56834
3,13821149
So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
2
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
1
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
3
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
1
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12
|
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
My question is NOT “why does the product topology satisfy the conditions of a categorical product in Top”.
Rather, my question is, why does it do so uniquely? I don’t require necessarily a fully rigorous proof, but an conceptual understanding.
My guess is: If we take the cartesian product between two topological spaces $X,Y$, but we endow this product with a trivial topology ${emptyset, Xtimes Y}$, then this also satisfies the categorical product conditions:
For any topological space $Z$ and continuous functions $f:Zto X, g:Zto Y$, there is a unique function $h=ftimes g:Zto Xtimes Y$. We don’t need to put any more conditions on the topology of $Xtimes Y$ for $h$ to be unique. (This is essentially because the cartesian product $Xtimes Y$ is already “set-theoretically restricted” enough to ensure uniqueness of $h$). Therefore, we can put any topology on $Xtimes Y$, so long as it makes $h$ unique for all such $(Z,f,g)$. One topology that does this is the product topology, but another one that does it is the topology ${emptyset,Xtimes Y }$.
What’s wrong with my argument?
EDIT: I Immediately realized that while this topology gets you a continuous $h$, it doesn’t get you continuous projections $pi_X,pi_Y$...
general-topology category-theory
asked Nov 21 at 12:21
user56834
3,13821149
My question is NOT “why does the product topology satisfy the conditions of a categorical product in Top”.
Rather, my question is, why does it do so uniquely? I don’t require necessarily a fully rigorous proof, but an conceptual understanding.
My guess is: If we take the cartesian product between two topological spaces $X,Y$, but we endow this product with a trivial topology ${emptyset, Xtimes Y}$, then this also satisfies the categorical product conditions:
For any topological space $Z$ and continuous functions $f:Zto X, g:Zto Y$, there is a unique function $h=ftimes g:Zto Xtimes Y$. We don’t need to put any more conditions on the topology of $Xtimes Y$ for $h$ to be unique. (This is essentially because the cartesian product $Xtimes Y$ is already “set-theoretically restricted” enough to ensure uniqueness of $h$). Therefore, we can put any topology on $Xtimes Y$, so long as it makes $h$ unique for all such $(Z,f,g)$. One topology that does this is the product topology, but another one that does it is the topology ${emptyset,Xtimes Y }$.
What’s wrong with my argument?
EDIT: I Immediately realized that while this topology gets you a continuous $h$, it doesn’t get you continuous projections $pi_X,pi_Y$...
general-topology category-theory
general-topology category-theory
asked Nov 21 at 12:21
user56834
3,13821149
asked Nov 21 at 12:21
user56834
3,13821149
asked Nov 21 at 12:21
user56834
3,13821149
asked Nov 21 at 12:21
user56834
3,13821149
asked Nov 21 at 12:21
user56834
3,13821149
3,13821149
So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
2
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
1
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
3
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
1
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12
|
show 1 more comment
So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
2
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
1
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
3
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
1
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12
So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
2
2
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
1
1
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
3
3
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
1
1
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12
|
show 1 more comment
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So basically you answered the question yourself. So what else do you need?
– freakish
Nov 21 at 12:25
2
@freakish, yes you’re right, lol. What should I do? Delete the question? What’s the standard practice?
– user56834
Nov 21 at 12:26
1
I guess deleting the question is fine.
– freakish
Nov 21 at 12:27
3
The general answer is that products are always determined up to isomorphism in their category.
– Kevin Carlson
Nov 21 at 15:54
1
More generally, limits are unique up to isomorphism, where they exist, and the product is an example of a limit. I second @MusaAl-hassy: you shouldn't delete the question, as it may be useful when others search similar questions. Instead, you can add an answer yourself to this!
– B. Mehta
Nov 22 at 17:12