Mixing
Defining a universal property
I was recently introduced to category theory and am looking to verify (proof?) a universal property from a diagram as shown.
In this example, the description says:
Take a topological space covered by two open subsets: $X = U cup V$. The diagram of inclusion maps has a universal property in the world of topological spaces and continuous maps.
Attempt at verifying a universal property:
Let $f: U → Y$ and
$g : V → Y$ be any continuous functions so that $f circ i$ = $g circ j$. Then there exists a unique continuous function $h : X → Y$ such that $h circ i' =g$ and $h circ j' = f$ so that the diagram commutes.
I'd appreciate if someone could guide me on the right track towards my first verification of universal property.
category-theory definition universal-property
asked Sep 21 '17 at 19:36
misheekoh
444318
add a comment |
I was recently introduced to category theory and am looking to verify (proof?) a universal property from a diagram as shown.
In this example, the description says:
Take a topological space covered by two open subsets: $X = U cup V$. The diagram of inclusion maps has a universal property in the world of topological spaces and continuous maps.
Attempt at verifying a universal property:
Let $f: U → Y$ and
$g : V → Y$ be any continuous functions so that $f circ i$ = $g circ j$. Then there exists a unique continuous function $h : X → Y$ such that $h circ i' =g$ and $h circ j' = f$ so that the diagram commutes.
I'd appreciate if someone could guide me on the right track towards my first verification of universal property.
category-theory definition universal-property
asked Sep 21 '17 at 19:36
misheekoh
444318
add a comment |
I was recently introduced to category theory and am looking to verify (proof?) a universal property from a diagram as shown.
In this example, the description says:
Take a topological space covered by two open subsets: $X = U cup V$. The diagram of inclusion maps has a universal property in the world of topological spaces and continuous maps.
Attempt at verifying a universal property:
Let $f: U → Y$ and
$g : V → Y$ be any continuous functions so that $f circ i$ = $g circ j$. Then there exists a unique continuous function $h : X → Y$ such that $h circ i' =g$ and $h circ j' = f$ so that the diagram commutes.
I'd appreciate if someone could guide me on the right track towards my first verification of universal property.
category-theory definition universal-property
asked Sep 21 '17 at 19:36
misheekoh
444318
I was recently introduced to category theory and am looking to verify (proof?) a universal property from a diagram as shown.
In this example, the description says:
Take a topological space covered by two open subsets: $X = U cup V$. The diagram of inclusion maps has a universal property in the world of topological spaces and continuous maps.
Attempt at verifying a universal property:
Let $f: U → Y$ and
$g : V → Y$ be any continuous functions so that $f circ i$ = $g circ j$. Then there exists a unique continuous function $h : X → Y$ such that $h circ i' =g$ and $h circ j' = f$ so that the diagram commutes.
I'd appreciate if someone could guide me on the right track towards my first verification of universal property.
category-theory definition universal-property
category-theory definition universal-property
asked Sep 21 '17 at 19:36
misheekoh
444318
asked Sep 21 '17 at 19:36
misheekoh
444318
asked Sep 21 '17 at 19:36
misheekoh
444318
asked Sep 21 '17 at 19:36
misheekoh
444318
asked Sep 21 '17 at 19:36
misheekoh
444318
444318
add a comment |
add a comment |
1 Answer
1
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oldest
votes
If you have continuous functions $f: Uto Y$ and $g: Vto Y$ you can define $h:Ucup Vto Y$ as follows
$$
h(x)=begin{cases}
f(x) & xin U\
g(x) & xin V
end{cases}
$$
I'll leave it to you to show that, given the conditions of your problem, $h$ is well-defined and continuous.
In the more general language of category theory, your diagram is a special case of a pushout (look here) in the category of topological spaces.
answered Sep 21 '17 at 19:53
Hamed
4,758521
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you have continuous functions $f: Uto Y$ and $g: Vto Y$ you can define $h:Ucup Vto Y$ as follows
$$
h(x)=begin{cases}
f(x) & xin U\
g(x) & xin V
end{cases}
$$
I'll leave it to you to show that, given the conditions of your problem, $h$ is well-defined and continuous.
In the more general language of category theory, your diagram is a special case of a pushout (look here) in the category of topological spaces.
answered Sep 21 '17 at 19:53
Hamed
4,758521
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
add a comment |
If you have continuous functions $f: Uto Y$ and $g: Vto Y$ you can define $h:Ucup Vto Y$ as follows
$$
h(x)=begin{cases}
f(x) & xin U\
g(x) & xin V
end{cases}
$$
I'll leave it to you to show that, given the conditions of your problem, $h$ is well-defined and continuous.
In the more general language of category theory, your diagram is a special case of a pushout (look here) in the category of topological spaces.
answered Sep 21 '17 at 19:53
Hamed
4,758521
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
add a comment |
If you have continuous functions $f: Uto Y$ and $g: Vto Y$ you can define $h:Ucup Vto Y$ as follows
$$
h(x)=begin{cases}
f(x) & xin U\
g(x) & xin V
end{cases}
$$
I'll leave it to you to show that, given the conditions of your problem, $h$ is well-defined and continuous.
In the more general language of category theory, your diagram is a special case of a pushout (look here) in the category of topological spaces.
answered Sep 21 '17 at 19:53
Hamed
4,758521
If you have continuous functions $f: Uto Y$ and $g: Vto Y$ you can define $h:Ucup Vto Y$ as follows
$$
h(x)=begin{cases}
f(x) & xin U\
g(x) & xin V
end{cases}
$$
I'll leave it to you to show that, given the conditions of your problem, $h$ is well-defined and continuous.
In the more general language of category theory, your diagram is a special case of a pushout (look here) in the category of topological spaces.
answered Sep 21 '17 at 19:53
Hamed
4,758521
edited Nov 20 '18 at 5:09
answered Sep 21 '17 at 19:53
Hamed
4,758521
answered Sep 21 '17 at 19:53
Hamed
4,758521
answered Sep 21 '17 at 19:53
Hamed
4,758521
4,758521
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
add a comment |
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Thanks for ur input! Much appreciated.
– misheekoh
Sep 21 '17 at 19:58
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
Could you define what u mean by "well-defined"
– misheekoh
Sep 21 '17 at 20:01
1
1
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
The way I defined $h$, it is not necessarily a function. Take a point $xin Ucap V$. Then where does $x$ go? $f(x)$ or $g(x)$? If $f(x)neq g(x)$ for some point $xin Ucap V$, then $h$ is not a function. You need to check, therefore, that this does not happen.
– Hamed
Sep 21 '17 at 20:02
add a comment |
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