Prop. A.17 in Hatcher
up vote
1
down vote
favorite
Here is the statement and proof in question.
I don't understand how they can say that $W$ has a quotient topolgy from $Xtimes Z$ and that there is a quotient map $g$. It seems like this what we are trying to prove. To me, a quotient map is a certain kind of surjective, continuous map. I also don't understand why we want to show that $h$ is continuous.
Also if someone could tell me what the subsript $c$ means here that would be greatly appreciated:
general-topology
asked Nov 23 at 15:44
SihOASHoihd
18312
add a comment |
up vote
1
down vote
favorite
Here is the statement and proof in question.
I don't understand how they can say that $W$ has a quotient topolgy from $Xtimes Z$ and that there is a quotient map $g$. It seems like this what we are trying to prove. To me, a quotient map is a certain kind of surjective, continuous map. I also don't understand why we want to show that $h$ is continuous.
Also if someone could tell me what the subsript $c$ means here that would be greatly appreciated:
general-topology
asked Nov 23 at 15:44
SihOASHoihd
18312
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is the statement and proof in question.
I don't understand how they can say that $W$ has a quotient topolgy from $Xtimes Z$ and that there is a quotient map $g$. It seems like this what we are trying to prove. To me, a quotient map is a certain kind of surjective, continuous map. I also don't understand why we want to show that $h$ is continuous.
Also if someone could tell me what the subsript $c$ means here that would be greatly appreciated:
general-topology
asked Nov 23 at 15:44
SihOASHoihd
18312
Here is the statement and proof in question.
I don't understand how they can say that $W$ has a quotient topolgy from $Xtimes Z$ and that there is a quotient map $g$. It seems like this what we are trying to prove. To me, a quotient map is a certain kind of surjective, continuous map. I also don't understand why we want to show that $h$ is continuous.
Also if someone could tell me what the subsript $c$ means here that would be greatly appreciated:
general-topology
general-topology
asked Nov 23 at 15:44
SihOASHoihd
18312
asked Nov 23 at 15:44
SihOASHoihd
18312
asked Nov 23 at 15:44
SihOASHoihd
18312
asked Nov 23 at 15:44
SihOASHoihd
18312
asked Nov 23 at 15:44
SihOASHoihd
18312
18312
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17
add a comment |
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y times Z$ as set and since he wants $ftimes 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A subset W$ to be open iff $(f times 1)^{-1}(A)$ is open in $X times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology
This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/sim$ with respect to $sim$ an equivalence relation.
Now we want to prove that the product topology on $Y times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h colon Y times Z rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is
https://ncatlab.org/nlab/show/compactly+generated+topological+space
basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C subset X$ will be compactly closed, that is for every continuous function $f colon K rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X times Y$ needs not to be a CW complex, while $(X times Y)_c$ admits the structure of CW complex.
answered Nov 23 at 16:14
N.B.
682313
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y times Z$ as set and since he wants $ftimes 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A subset W$ to be open iff $(f times 1)^{-1}(A)$ is open in $X times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology
This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/sim$ with respect to $sim$ an equivalence relation.
Now we want to prove that the product topology on $Y times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h colon Y times Z rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is
https://ncatlab.org/nlab/show/compactly+generated+topological+space
basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C subset X$ will be compactly closed, that is for every continuous function $f colon K rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X times Y$ needs not to be a CW complex, while $(X times Y)_c$ admits the structure of CW complex.
answered Nov 23 at 16:14
N.B.
682313
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
add a comment |
up vote
1
down vote
accepted
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y times Z$ as set and since he wants $ftimes 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A subset W$ to be open iff $(f times 1)^{-1}(A)$ is open in $X times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology
This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/sim$ with respect to $sim$ an equivalence relation.
Now we want to prove that the product topology on $Y times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h colon Y times Z rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is
https://ncatlab.org/nlab/show/compactly+generated+topological+space
basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C subset X$ will be compactly closed, that is for every continuous function $f colon K rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X times Y$ needs not to be a CW complex, while $(X times Y)_c$ admits the structure of CW complex.
answered Nov 23 at 16:14
N.B.
682313
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y times Z$ as set and since he wants $ftimes 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A subset W$ to be open iff $(f times 1)^{-1}(A)$ is open in $X times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology
This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/sim$ with respect to $sim$ an equivalence relation.
Now we want to prove that the product topology on $Y times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h colon Y times Z rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is
https://ncatlab.org/nlab/show/compactly+generated+topological+space
basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C subset X$ will be compactly closed, that is for every continuous function $f colon K rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X times Y$ needs not to be a CW complex, while $(X times Y)_c$ admits the structure of CW complex.
answered Nov 23 at 16:14
N.B.
682313
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y times Z$ as set and since he wants $ftimes 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A subset W$ to be open iff $(f times 1)^{-1}(A)$ is open in $X times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology
This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/sim$ with respect to $sim$ an equivalence relation.
Now we want to prove that the product topology on $Y times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h colon Y times Z rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is
https://ncatlab.org/nlab/show/compactly+generated+topological+space
basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C subset X$ will be compactly closed, that is for every continuous function $f colon K rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X times Y$ needs not to be a CW complex, while $(X times Y)_c$ admits the structure of CW complex.
answered Nov 23 at 16:14
N.B.
682313
edited Nov 23 at 16:18
answered Nov 23 at 16:14
N.B.
682313
answered Nov 23 at 16:14
N.B.
682313
answered Nov 23 at 16:14
N.B.
682313
682313
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
add a comment |
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
Yes, the context is compactly generated spaces, so that checks out. But he just starts using the notation without mentioning anything... Thanks.
– SihOASHoihd
Nov 23 at 16:17
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010508%2fprop-a-17-in-hatcher%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I guess $g$ is $f times mathbf{1}$ considered as a set map and then we use this map to define a quotient topology on $Y times Z$?
– SihOASHoihd
Nov 23 at 15:50
I think I get it. $W$ is $Y times Z$ with the quotient topology induced by the surjective map $f times mathbf{1}$ and $Ytimes Z$ has the product topology. I guess I have to used to Hatcher's wording. I am still interested in the subscript c though.
– SihOASHoihd
Nov 23 at 16:09
Yes, $W$ is the set $Y times Z$ but with the final topology induced by the map $f times 1$! This is not necessarily $Y times Z$ which is the space given by the product topology. The point of the proof is to show that they coincide as spaces, that is the two topologies coincide. As sets they are equal.
– N.B.
Nov 23 at 16:17