Can I use Seifert-van Kampen theorem infinite times The Next CEO of Stack OverflowPerturbation trick in the proof of Seifert-van-KampenSeifert-van-Kampen and free product with amalgamationFundamental group via Van KampenHow to use Seifert–van Kampen to find $pi_1( RP^n ∨ RP^n)$Practicing Seifert van KampenRealising Seifert-van Kampen in 2-complexesComputing fundamental group with Seifert-van Kampen theoremChoosing Open Sets in Applying the Seifert-Van Kampen Theorem to Topological space$H_1(F)$ for an orientable, compact, connected surfacehow to calculate the fundamental group using van kampen theorem?
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Can I use Seifert-van Kampen theorem infinite times
The Next CEO of Stack OverflowPerturbation trick in the proof of Seifert-van-KampenSeifert-van-Kampen and free product with amalgamationFundamental group via Van KampenHow to use Seifert–van Kampen to find $pi_1( RP^n ∨ RP^n)$Practicing Seifert van KampenRealising Seifert-van Kampen in 2-complexesComputing fundamental group with Seifert-van Kampen theoremChoosing Open Sets in Applying the Seifert-Van Kampen Theorem to Topological space$H_1(F)$ for an orientable, compact, connected surfacehow to calculate the fundamental group using van kampen theorem?
$begingroup$
I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?
Thanks.
general-topology algebraic-topology fundamental-groups
$endgroup$
add a comment |
$begingroup$
I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?
Thanks.
general-topology algebraic-topology fundamental-groups
$endgroup$
add a comment |
$begingroup$
I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?
Thanks.
general-topology algebraic-topology fundamental-groups
$endgroup$
I know the definition of Seifert-van Kampen theorem for a topological space "made" with 2 parts. Is not difficult to see that if I use the theorem a finite number of times to calculate a the fundamental group of a topological space made from finite many parts, it is valid. But, can I use the theorem infinite times? For example, to show that te fundamental group of a orientable surface of infinite genus is isomorphic to the free group with infinite generators?
Thanks.
general-topology algebraic-topology fundamental-groups
general-topology algebraic-topology fundamental-groups
asked Mar 24 at 4:48
J.RodriguezJ.Rodriguez
19710
19710
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.
Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
$$requireAMScd
beginCD
pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
@VVV @VVV \
pi_1(X, ast) @= pi_1(X,ast) @= ldots
endCD$$
which gives a map from the direct limit
$$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$
Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.
Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.
In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.
While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
$$ varinjlim F_n = F_infty,$$
where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.
$endgroup$
add a comment |
$begingroup$
As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.
Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.
Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.
$endgroup$
add a comment |
$begingroup$
What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.
In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.
Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.
Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
$$requireAMScd
beginCD
pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
@VVV @VVV \
pi_1(X, ast) @= pi_1(X,ast) @= ldots
endCD$$
which gives a map from the direct limit
$$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$
Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.
Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.
In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.
While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
$$ varinjlim F_n = F_infty,$$
where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.
$endgroup$
add a comment |
$begingroup$
If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.
Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
$$requireAMScd
beginCD
pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
@VVV @VVV \
pi_1(X, ast) @= pi_1(X,ast) @= ldots
endCD$$
which gives a map from the direct limit
$$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$
Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.
Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.
In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.
While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
$$ varinjlim F_n = F_infty,$$
where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.
$endgroup$
add a comment |
$begingroup$
If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.
Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
$$requireAMScd
beginCD
pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
@VVV @VVV \
pi_1(X, ast) @= pi_1(X,ast) @= ldots
endCD$$
which gives a map from the direct limit
$$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$
Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.
Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.
In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.
While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
$$ varinjlim F_n = F_infty,$$
where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.
$endgroup$
If we have an infinite cover of a space $X$, one approach is to compute $pi_1$ of finite unions of those covers, thus reducing the problem to the case of an increasing union $U_1 subseteq U_2 subseteq cdots subseteq X = bigcup_i=1^infty U_i$. This is handled via a direct limit of the fundamental groups of the open sets $U_i$.
Given a space $X$ which is an increasing union of open sets $U_1 subseteq U_2 subseteq ldots$, with a specified basepoint $ast in U_1$, we have a diagram
$$requireAMScd
beginCD
pi_1(U_1,ast) @> >> pi_1(U_2,ast) @>>> ldots \
@VVV @VVV \
pi_1(X, ast) @= pi_1(X,ast) @= ldots
endCD$$
which gives a map from the direct limit
$$Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast).$$
Claim: $Phi: varinjlim pi_1(U_i, ast) to pi_1(X,ast)$ is an isomorphism.
Proof: To see that $Phi$ is surjective, observe that if $gamma: (S^1, ast) to (X,ast)$ is a based loop, then $gamma(S^1) subseteq cup_i=1^infty U_i$ is compact, so its image is contained in a finite subcover of $U_i_i=1^infty$ and thus in $U_i$ for some $i$.
In the wise words of one of my topology professors, "injectivity is just surjectivity, one dimension higher." If $[gamma_1]$ and $[gamma_2]$ are homotopy classes in the direct limit which are (based) homotopic in $X$ via $H: S^1 times I to X$, then the image of $H$ is (by the same argument) contained in $U_j$ for some $j$, which insures that $[gamma_1] = [gamma_2]$ in $pi_1(U_k,ast)$ for $k geq j$. This shows injectivity.
While a direct limit of a system of groups may be intimidating, it allows us to compute. For instance, if $F_n$ is the free group on generators $x_1,x_2,ldots, x_n$, with maps $F_n to F_n+1$ given by including the generators $x_1,x_2,ldots, x_n subseteq x_1,x_2,ldots, x_n+1$, then
$$ varinjlim F_n = F_infty,$$
where $F_infty$ is the free group on infinitely many generators $x_1, x_2,ldots$.
answered Mar 24 at 5:31
Joshua MundingerJoshua Mundinger
2,9071028
2,9071028
add a comment |
add a comment |
$begingroup$
As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.
Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.
Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.
$endgroup$
add a comment |
$begingroup$
As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.
Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.
Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.
$endgroup$
add a comment |
$begingroup$
As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.
Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.
Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.
$endgroup$
As Joshua Mundinger has nicely explained, you can "iterate" the Seifert-van Kampen theorem infinitely many times by taking a direct limit, using the fact that $pi_1$ preserves direct limits of open inclusions. However, there is also a more direct way to use Seifert-van Kampen for infinite open covers: there is a version of the theorem that applies to covers by an arbitrary number of open sets, rather than just two open sets.
Here's one version of the statement (this is Theorem 1.20 in Hatcher's Algebraic Topology, for instance). Suppose $(X,*)$ is a pointed space and $(U_i)_iin I$ is an open cover of $X$ such that $*in U_i$ for all $i$. Suppose furthermore that $U_i$, $U_icap U_j$, and $U_icap U_jcap U_k$ are path-connected for all $i,j,kin I$. Then $pi_1(X,*)$ is isomorphic (via the obvious map) to the quotient of the free product of the groups $pi_1(U_i,*)$ by relations which say the two maps $pi_1(U_icap U_j,*)to pi_1(U_i,*)$ and $pi_1(U_icap U_j,*)topi_1(U_j,*)$ become equal.
Note that this is useful not just for infinite covers, but also for finite covers by more than two sets, allowing you to compute the final result all at once rather than needing to iterate.
answered Mar 24 at 6:22
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.
In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.
Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.
$endgroup$
add a comment |
$begingroup$
What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.
In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.
Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.
$endgroup$
add a comment |
$begingroup$
What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.
In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.
Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.
$endgroup$
What would it even mean to apply the theorem infinitely many times? That would have to be some sort of inductive process - and you have to be very careful about the sorts of infinities you get out of that.
In this case, adding one hole at a time, we get as subgroups a chain of free groups on increasing numbers of generators - but since there are still infinitely many holes at any stage of the process, there will be another factor that we simply can't cleanly measure this way.
Now, we can take the direct limit of those free subgroups we found to say that there's an infinite free group that's a subgroup of the fundamental group. We can't say more without additional tools.
answered Mar 24 at 5:27
jmerryjmerry
16.9k11633
16.9k11633
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