Mixing
About the universal property of initial and final topologies
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I have recently seen in my topology course that if $X$ is any set,
Given a family of functions $f_i : X to Y_i_i in I$, with $(Y_i, tau_i)_i$ topological spaces, the initial topology on $X$ with respect to this family is the one generated by $f_i^-1(U_i) : i in I, U_i in tau_i $, which is the coarsest such that the functions $f_i$ are continuous, and verifies $h : Z to X$ continuous if and only if $f_ih$ is continuous for all $i$ in $I$.
In a similar way, given a family $f_i : Y_i to X_i in I$, the final topology $tau$ in $X$ is defined by $U in tau $ if and only if $f_i^-1(U) in tau_i (forall i in I)$. This is the finest topology such that the family is countinuous, and $h : X to Z$ is continuous if and only if $hf_i$ is for all $i in I$.
It was also mentioned that the reciprocal of both statements is true, i.e. that if a topological space $(X, tau)$ verifies $h : Z to X$ (resp. $h:X to Z$) continuous if and only if $f_ih$ (resp. $hf_i$) continuous for all $i$, for all $h$, it has the initial (resp. final) topology with respect to the family $(f_i)_i$.
It is easy to see taking $h equiv id $ that the given topology $tau$ is finer/coarser than the initial/final topology, because all the functions $f_i$ are continuous.
Any hints on how to prove the other inclusion?
general-topology
asked 4 hours ago
Guido A.
4,708727
 |Â
show 9 more comments
up vote
3
down vote
favorite
I have recently seen in my topology course that if $X$ is any set,
Given a family of functions $f_i : X to Y_i_i in I$, with $(Y_i, tau_i)_i$ topological spaces, the initial topology on $X$ with respect to this family is the one generated by $f_i^-1(U_i) : i in I, U_i in tau_i $, which is the coarsest such that the functions $f_i$ are continuous, and verifies $h : Z to X$ continuous if and only if $f_ih$ is continuous for all $i$ in $I$.
In a similar way, given a family $f_i : Y_i to X_i in I$, the final topology $tau$ in $X$ is defined by $U in tau $ if and only if $f_i^-1(U) in tau_i (forall i in I)$. This is the finest topology such that the family is countinuous, and $h : X to Z$ is continuous if and only if $hf_i$ is for all $i in I$.
It was also mentioned that the reciprocal of both statements is true, i.e. that if a topological space $(X, tau)$ verifies $h : Z to X$ (resp. $h:X to Z$) continuous if and only if $f_ih$ (resp. $hf_i$) continuous for all $i$, for all $h$, it has the initial (resp. final) topology with respect to the family $(f_i)_i$.
It is easy to see taking $h equiv id $ that the given topology $tau$ is finer/coarser than the initial/final topology, because all the functions $f_i$ are continuous.
Any hints on how to prove the other inclusion?
general-topology
asked 4 hours ago
Guido A.
4,708727
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
1
Ok, now it makes sense.
– celtschk
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
1
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago
 |Â
show 9 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have recently seen in my topology course that if $X$ is any set,
Given a family of functions $f_i : X to Y_i_i in I$, with $(Y_i, tau_i)_i$ topological spaces, the initial topology on $X$ with respect to this family is the one generated by $f_i^-1(U_i) : i in I, U_i in tau_i $, which is the coarsest such that the functions $f_i$ are continuous, and verifies $h : Z to X$ continuous if and only if $f_ih$ is continuous for all $i$ in $I$.
In a similar way, given a family $f_i : Y_i to X_i in I$, the final topology $tau$ in $X$ is defined by $U in tau $ if and only if $f_i^-1(U) in tau_i (forall i in I)$. This is the finest topology such that the family is countinuous, and $h : X to Z$ is continuous if and only if $hf_i$ is for all $i in I$.
It was also mentioned that the reciprocal of both statements is true, i.e. that if a topological space $(X, tau)$ verifies $h : Z to X$ (resp. $h:X to Z$) continuous if and only if $f_ih$ (resp. $hf_i$) continuous for all $i$, for all $h$, it has the initial (resp. final) topology with respect to the family $(f_i)_i$.
It is easy to see taking $h equiv id $ that the given topology $tau$ is finer/coarser than the initial/final topology, because all the functions $f_i$ are continuous.
Any hints on how to prove the other inclusion?
general-topology
asked 4 hours ago
Guido A.
4,708727
I have recently seen in my topology course that if $X$ is any set,
Given a family of functions $f_i : X to Y_i_i in I$, with $(Y_i, tau_i)_i$ topological spaces, the initial topology on $X$ with respect to this family is the one generated by $f_i^-1(U_i) : i in I, U_i in tau_i $, which is the coarsest such that the functions $f_i$ are continuous, and verifies $h : Z to X$ continuous if and only if $f_ih$ is continuous for all $i$ in $I$.
In a similar way, given a family $f_i : Y_i to X_i in I$, the final topology $tau$ in $X$ is defined by $U in tau $ if and only if $f_i^-1(U) in tau_i (forall i in I)$. This is the finest topology such that the family is countinuous, and $h : X to Z$ is continuous if and only if $hf_i$ is for all $i in I$.
It was also mentioned that the reciprocal of both statements is true, i.e. that if a topological space $(X, tau)$ verifies $h : Z to X$ (resp. $h:X to Z$) continuous if and only if $f_ih$ (resp. $hf_i$) continuous for all $i$, for all $h$, it has the initial (resp. final) topology with respect to the family $(f_i)_i$.
It is easy to see taking $h equiv id $ that the given topology $tau$ is finer/coarser than the initial/final topology, because all the functions $f_i$ are continuous.
Any hints on how to prove the other inclusion?
general-topology
general-topology
asked 4 hours ago
Guido A.
4,708727
asked 4 hours ago
Guido A.
4,708727
edited 2 hours ago
asked 4 hours ago
Guido A.
4,708727
asked 4 hours ago
Guido A.
4,708727
asked 4 hours ago
Guido A.
4,708727
4,708727
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
1
Ok, now it makes sense.
– celtschk
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
1
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago
 |Â
show 9 more comments
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
1
Ok, now it makes sense.
– celtschk
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
1
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
1
1
Ok, now it makes sense.
– celtschk
3 hours ago
Ok, now it makes sense.
– celtschk
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
1
1
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago
 |Â
show 9 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let it be that $tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Zto X$ is continuous iff $f_ih:Zto X_i$ for every $iin I$.
Further let it be that $tau'$ is the topology generated by the collection $mathcal V=f_i^-1(U_i)mid iin I, U_iintau_i$.
Then for every space $(Z,tau_Z)$ and every function $h:Zto X$ the following statements are equivalent:
- $h:(Z,tau_Z)to (X,tau')$ is continuous
- $h^-1(f_i^-1(U_i))subseteqtau_Z$ for every $iin I$ and every $U_iintau_i$
$f_ih:(Z,tau_Z)to (X,tau_i)$ is continuous for every $iin I$
$h:(Z,tau_Z)to (X,tau)$ is continuous.
Applying this on $mathsfid:(X,tau)to(X,tau')$ and $mathsfid:(X,tau')to(X,tau)$ we find that $tau=tau'$
answered 2 hours ago
drhab
88.9k541122
add a comment |Â
up vote
3
down vote
As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.
In the case of $tau$ being the initial topology, let $tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, tau')$ makes $f_i : (X,tau') rightarrow Y_i$ continuous for all $i$ and so $tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $f_i_i$. To see that $tau' subseteq tau$, it is equivalent to prove that $(X,tau) xrightarrowid (X,tau')$ is continuous. Equivalently, we can see that each composition
$$
(X,tau) xrightarrowid (X,tau') xrightarrowf_i Y_i
$$
is continuous, but these are the mappings $f_i : (X,tau) rightarrow Y_i$ which by definition of $tau$ ought to be continuous, thus proving the original claim.
The case for the final topology follows likewise.
answered 2 hours ago
Guido A.
4,708727
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let it be that $tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Zto X$ is continuous iff $f_ih:Zto X_i$ for every $iin I$.
Further let it be that $tau'$ is the topology generated by the collection $mathcal V=f_i^-1(U_i)mid iin I, U_iintau_i$.
Then for every space $(Z,tau_Z)$ and every function $h:Zto X$ the following statements are equivalent:
- $h:(Z,tau_Z)to (X,tau')$ is continuous
- $h^-1(f_i^-1(U_i))subseteqtau_Z$ for every $iin I$ and every $U_iintau_i$
$f_ih:(Z,tau_Z)to (X,tau_i)$ is continuous for every $iin I$
$h:(Z,tau_Z)to (X,tau)$ is continuous.
Applying this on $mathsfid:(X,tau)to(X,tau')$ and $mathsfid:(X,tau')to(X,tau)$ we find that $tau=tau'$
answered 2 hours ago
drhab
88.9k541122
add a comment |Â
up vote
2
down vote
accepted
Let it be that $tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Zto X$ is continuous iff $f_ih:Zto X_i$ for every $iin I$.
Further let it be that $tau'$ is the topology generated by the collection $mathcal V=f_i^-1(U_i)mid iin I, U_iintau_i$.
Then for every space $(Z,tau_Z)$ and every function $h:Zto X$ the following statements are equivalent:
- $h:(Z,tau_Z)to (X,tau')$ is continuous
- $h^-1(f_i^-1(U_i))subseteqtau_Z$ for every $iin I$ and every $U_iintau_i$
$f_ih:(Z,tau_Z)to (X,tau_i)$ is continuous for every $iin I$
$h:(Z,tau_Z)to (X,tau)$ is continuous.
Applying this on $mathsfid:(X,tau)to(X,tau')$ and $mathsfid:(X,tau')to(X,tau)$ we find that $tau=tau'$
answered 2 hours ago
drhab
88.9k541122
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let it be that $tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Zto X$ is continuous iff $f_ih:Zto X_i$ for every $iin I$.
Further let it be that $tau'$ is the topology generated by the collection $mathcal V=f_i^-1(U_i)mid iin I, U_iintau_i$.
Then for every space $(Z,tau_Z)$ and every function $h:Zto X$ the following statements are equivalent:
- $h:(Z,tau_Z)to (X,tau')$ is continuous
- $h^-1(f_i^-1(U_i))subseteqtau_Z$ for every $iin I$ and every $U_iintau_i$
$f_ih:(Z,tau_Z)to (X,tau_i)$ is continuous for every $iin I$
$h:(Z,tau_Z)to (X,tau)$ is continuous.
Applying this on $mathsfid:(X,tau)to(X,tau')$ and $mathsfid:(X,tau')to(X,tau)$ we find that $tau=tau'$
answered 2 hours ago
drhab
88.9k541122
Let it be that $tau$ is the topology on $X$ initialized by the $f_i$ in the sense that a function $h:Zto X$ is continuous iff $f_ih:Zto X_i$ for every $iin I$.
Further let it be that $tau'$ is the topology generated by the collection $mathcal V=f_i^-1(U_i)mid iin I, U_iintau_i$.
Then for every space $(Z,tau_Z)$ and every function $h:Zto X$ the following statements are equivalent:
- $h:(Z,tau_Z)to (X,tau')$ is continuous
- $h^-1(f_i^-1(U_i))subseteqtau_Z$ for every $iin I$ and every $U_iintau_i$
$f_ih:(Z,tau_Z)to (X,tau_i)$ is continuous for every $iin I$
$h:(Z,tau_Z)to (X,tau)$ is continuous.
Applying this on $mathsfid:(X,tau)to(X,tau')$ and $mathsfid:(X,tau')to(X,tau)$ we find that $tau=tau'$
answered 2 hours ago
drhab
88.9k541122
answered 2 hours ago
drhab
88.9k541122
answered 2 hours ago
drhab
88.9k541122
answered 2 hours ago
drhab
88.9k541122
88.9k541122
add a comment |Â
add a comment |Â
up vote
3
down vote
As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.
In the case of $tau$ being the initial topology, let $tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, tau')$ makes $f_i : (X,tau') rightarrow Y_i$ continuous for all $i$ and so $tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $f_i_i$. To see that $tau' subseteq tau$, it is equivalent to prove that $(X,tau) xrightarrowid (X,tau')$ is continuous. Equivalently, we can see that each composition
$$
(X,tau) xrightarrowid (X,tau') xrightarrowf_i Y_i
$$
is continuous, but these are the mappings $f_i : (X,tau) rightarrow Y_i$ which by definition of $tau$ ought to be continuous, thus proving the original claim.
The case for the final topology follows likewise.
answered 2 hours ago
Guido A.
4,708727
add a comment |Â
up vote
3
down vote
As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.
In the case of $tau$ being the initial topology, let $tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, tau')$ makes $f_i : (X,tau') rightarrow Y_i$ continuous for all $i$ and so $tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $f_i_i$. To see that $tau' subseteq tau$, it is equivalent to prove that $(X,tau) xrightarrowid (X,tau')$ is continuous. Equivalently, we can see that each composition
$$
(X,tau) xrightarrowid (X,tau') xrightarrowf_i Y_i
$$
is continuous, but these are the mappings $f_i : (X,tau) rightarrow Y_i$ which by definition of $tau$ ought to be continuous, thus proving the original claim.
The case for the final topology follows likewise.
answered 2 hours ago
Guido A.
4,708727
add a comment |Â
up vote
3
down vote
up vote
3
down vote
As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.
In the case of $tau$ being the initial topology, let $tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, tau')$ makes $f_i : (X,tau') rightarrow Y_i$ continuous for all $i$ and so $tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $f_i_i$. To see that $tau' subseteq tau$, it is equivalent to prove that $(X,tau) xrightarrowid (X,tau')$ is continuous. Equivalently, we can see that each composition
$$
(X,tau) xrightarrowid (X,tau') xrightarrowf_i Y_i
$$
is continuous, but these are the mappings $f_i : (X,tau) rightarrow Y_i$ which by definition of $tau$ ought to be continuous, thus proving the original claim.
The case for the final topology follows likewise.
answered 2 hours ago
Guido A.
4,708727
As referenced by Theoretical Economist, an answer by Henno Brandsma indicates that the other inclusion can be proved by considering the identity as well, but changing the topologies in the domain/codomain.
In the case of $tau$ being the initial topology, let $tau'$ be a topology so that for any function $h$ we have that $h$ continuous if and only if $f_ih$ is continuous for all $i$. As I stated in the post it is easy to see that any other topology satisfying this is finer than the initial topology: the identity on $(X, tau')$ makes $f_i : (X,tau') rightarrow Y_i$ continuous for all $i$ and so $tau'$ contains the inital topology, generated by preimages of open sets of the $Y_i$ via $f_i_i$. To see that $tau' subseteq tau$, it is equivalent to prove that $(X,tau) xrightarrowid (X,tau')$ is continuous. Equivalently, we can see that each composition
$$
(X,tau) xrightarrowid (X,tau') xrightarrowf_i Y_i
$$
is continuous, but these are the mappings $f_i : (X,tau) rightarrow Y_i$ which by definition of $tau$ ought to be continuous, thus proving the original claim.
The case for the final topology follows likewise.
answered 2 hours ago
Guido A.
4,708727
answered 2 hours ago
Guido A.
4,708727
answered 2 hours ago
Guido A.
4,708727
answered 2 hours ago
Guido A.
4,708727
4,708727
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2917657%2fabout-the-universal-property-of-initial-and-final-topologies%23new-answer', 'question_page');
);
Post as a guest
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
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
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
@celtschk I totally had a typo there, I've fixed it.
– Guido A.
3 hours ago
1
Ok, now it makes sense.
– celtschk
3 hours ago
"which is the coarser so that the functions" -- this seems like a typo?
– Theoretical Economist
3 hours ago
@TheoreticalEconomist gramatically or conceptually? I tried to say that this is the topology with 'less open sets' so that every function in the family is continuous.
– Guido A.
3 hours ago
1
Grammatically. I think you mean “which is the coarsest topology such that the functionsâ€�
– Theoretical Economist
3 hours ago