Is it true that if inclusion of a boundary component is a homotopy equivalence then the manifold deformation retracts onto it?Is it possible for a closed manifold to deformation retract onto a proper subset of itself?Homotopy equivalence iff both spaces are deformation retractshomotopy type of the manifold minus the boundaryIs it possible for a closed manifold to deformation retract onto a proper subset of itself?Homework: The Klein bottle retracts onto a loop.Show that $X$ deformation retracts to any point in the segment $[0,1]times lbrace 0 rbrace$, but not to any other point.Does the homotopy equivalence of a space with its subspace imply deformation retract?How to show if $A subset V subset X$ and $V$ deformation retracts onto $A$, then $H_n(X,A) simeq H_n(X,V)$.Examples for a contractible pair that induce a non-surjective fundamental groups homomorphism.Does there exist a compact manifold with boundary $M$ that is contractible but does not (strong) deformation retract onto a point?If $Sigma$ is a homotopy sphere, then $Sigma#(-Sigma)$ bounds a contractible manifold.

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Is it true that if inclusion of a boundary component is a homotopy equivalence then the manifold deformation retracts onto it?


Is it possible for a closed manifold to deformation retract onto a proper subset of itself?Homotopy equivalence iff both spaces are deformation retractshomotopy type of the manifold minus the boundaryIs it possible for a closed manifold to deformation retract onto a proper subset of itself?Homework: The Klein bottle retracts onto a loop.Show that $X$ deformation retracts to any point in the segment $[0,1]times lbrace 0 rbrace$, but not to any other point.Does the homotopy equivalence of a space with its subspace imply deformation retract?How to show if $A subset V subset X$ and $V$ deformation retracts onto $A$, then $H_n(X,A) simeq H_n(X,V)$.Examples for a contractible pair that induce a non-surjective fundamental groups homomorphism.Does there exist a compact manifold with boundary $M$ that is contractible but does not (strong) deformation retract onto a point?If $Sigma$ is a homotopy sphere, then $Sigma#(-Sigma)$ bounds a contractible manifold.













8












$begingroup$


In topological spaces, if $Asubset X$ and the inclusion is a homotopy equivalence, that doesn't imply that $X$ deformation retracts onto $A$. For example take one of those examples of a contractible space that doesn't deformation retract to a point.



Even in manifolds, a closed manifold doesn't deformation retract to any proper subset, one can show that with homology as done here by Olivier Bégassat.



So my question is: in the case of a manifold with boundary $(M,partial M)$, with $Nsubset partial M$ a boundary component such that $Nhookrightarrow M$ is a homotopy equivalence, is it true that $M$ deformation retracts onto $N$?



Feel free to assume more hypothesis if needed.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
    $endgroup$
    – ThorbenK
    Mar 11 at 12:42
















8












$begingroup$


In topological spaces, if $Asubset X$ and the inclusion is a homotopy equivalence, that doesn't imply that $X$ deformation retracts onto $A$. For example take one of those examples of a contractible space that doesn't deformation retract to a point.



Even in manifolds, a closed manifold doesn't deformation retract to any proper subset, one can show that with homology as done here by Olivier Bégassat.



So my question is: in the case of a manifold with boundary $(M,partial M)$, with $Nsubset partial M$ a boundary component such that $Nhookrightarrow M$ is a homotopy equivalence, is it true that $M$ deformation retracts onto $N$?



Feel free to assume more hypothesis if needed.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
    $endgroup$
    – ThorbenK
    Mar 11 at 12:42














8












8








8


3



$begingroup$


In topological spaces, if $Asubset X$ and the inclusion is a homotopy equivalence, that doesn't imply that $X$ deformation retracts onto $A$. For example take one of those examples of a contractible space that doesn't deformation retract to a point.



Even in manifolds, a closed manifold doesn't deformation retract to any proper subset, one can show that with homology as done here by Olivier Bégassat.



So my question is: in the case of a manifold with boundary $(M,partial M)$, with $Nsubset partial M$ a boundary component such that $Nhookrightarrow M$ is a homotopy equivalence, is it true that $M$ deformation retracts onto $N$?



Feel free to assume more hypothesis if needed.










share|cite|improve this question











$endgroup$




In topological spaces, if $Asubset X$ and the inclusion is a homotopy equivalence, that doesn't imply that $X$ deformation retracts onto $A$. For example take one of those examples of a contractible space that doesn't deformation retract to a point.



Even in manifolds, a closed manifold doesn't deformation retract to any proper subset, one can show that with homology as done here by Olivier Bégassat.



So my question is: in the case of a manifold with boundary $(M,partial M)$, with $Nsubset partial M$ a boundary component such that $Nhookrightarrow M$ is a homotopy equivalence, is it true that $M$ deformation retracts onto $N$?



Feel free to assume more hypothesis if needed.







general-topology geometry differential-geometry algebraic-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







giulio

















asked Mar 11 at 8:55









giuliogiulio

473413




473413







  • 1




    $begingroup$
    If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
    $endgroup$
    – ThorbenK
    Mar 11 at 12:42













  • 1




    $begingroup$
    If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
    $endgroup$
    – ThorbenK
    Mar 11 at 12:42








1




1




$begingroup$
If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
$endgroup$
– ThorbenK
Mar 11 at 12:42





$begingroup$
If everything in question is simply connected and of high enough dimension than it is even isomorphic to a cylinder by poincare duality and the h-cobordism theorem. I guess the S-cobordism theorem is a good place to look for counterexamples for non-simply connected things.
$endgroup$
– ThorbenK
Mar 11 at 12:42











1 Answer
1






active

oldest

votes


















8












$begingroup$

Yes, you get a deformation retract. More formally...




Suppose $M$ is a topological manifold, $Nsubseteq partial M subseteq M$ is a component of $partial M$ and that the inclusion $Nrightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.




Here's the idea of the proof:



From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.



Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.



So, some closed neighborhood of $N$ in $M$ is of the form $Ntimes [0,1] = Ntimes I$, where we are identifying $N$ with $Ntimes 0subseteq M$..



Now suppose $f:Mrightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:Ntimes Irightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:Mtimes Irightarrow X$ satisfying two conditions. First, for any $nin N$, $tin I$, that $G(n,t) = F(n,t)$. Second, for any $min M$, $G(m,1) = f(m)$.



To that end, set $G(m,t) = begincases f(m) & mnotin Ntimes [0,1]\ F(n, maxs,t) & m=(n,s)in Ntimes [0,1].endcases$



To check the first condition, note that $nin Ncong Ntimes 0$ means the corresonding $s$ is $s= 0$. Since $tin[0,1]$, $maxs,t = max0,t= t$, so $G(n,t) = F(n,t).$



To check the second condition, we have $G(m,t) = f(m)$ if $mnotin Ntimes [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)in Ntimes [0,1]$, then because we are taking $t=1$, we have $maxs,t = t$. So $G(m,1) = F(m,1) = f(m)$.



Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.



But, if $min Ntimes [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $maxs,t = s=1$, so $G(m) = F(n, maxs,t) = F(n,1) = f(n).$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 for pointing out that you just need to verify the homotopy extension property.
    $endgroup$
    – William
    Mar 11 at 17:35










  • $begingroup$
    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
    $endgroup$
    – Moishe Kohan
    2 days ago










  • $begingroup$
    Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
    $endgroup$
    – Jason DeVito
    yesterday










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1 Answer
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1 Answer
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active

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active

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active

oldest

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8












$begingroup$

Yes, you get a deformation retract. More formally...




Suppose $M$ is a topological manifold, $Nsubseteq partial M subseteq M$ is a component of $partial M$ and that the inclusion $Nrightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.




Here's the idea of the proof:



From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.



Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.



So, some closed neighborhood of $N$ in $M$ is of the form $Ntimes [0,1] = Ntimes I$, where we are identifying $N$ with $Ntimes 0subseteq M$..



Now suppose $f:Mrightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:Ntimes Irightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:Mtimes Irightarrow X$ satisfying two conditions. First, for any $nin N$, $tin I$, that $G(n,t) = F(n,t)$. Second, for any $min M$, $G(m,1) = f(m)$.



To that end, set $G(m,t) = begincases f(m) & mnotin Ntimes [0,1]\ F(n, maxs,t) & m=(n,s)in Ntimes [0,1].endcases$



To check the first condition, note that $nin Ncong Ntimes 0$ means the corresonding $s$ is $s= 0$. Since $tin[0,1]$, $maxs,t = max0,t= t$, so $G(n,t) = F(n,t).$



To check the second condition, we have $G(m,t) = f(m)$ if $mnotin Ntimes [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)in Ntimes [0,1]$, then because we are taking $t=1$, we have $maxs,t = t$. So $G(m,1) = F(m,1) = f(m)$.



Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.



But, if $min Ntimes [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $maxs,t = s=1$, so $G(m) = F(n, maxs,t) = F(n,1) = f(n).$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 for pointing out that you just need to verify the homotopy extension property.
    $endgroup$
    – William
    Mar 11 at 17:35










  • $begingroup$
    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
    $endgroup$
    – Moishe Kohan
    2 days ago










  • $begingroup$
    Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
    $endgroup$
    – Jason DeVito
    yesterday















8












$begingroup$

Yes, you get a deformation retract. More formally...




Suppose $M$ is a topological manifold, $Nsubseteq partial M subseteq M$ is a component of $partial M$ and that the inclusion $Nrightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.




Here's the idea of the proof:



From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.



Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.



So, some closed neighborhood of $N$ in $M$ is of the form $Ntimes [0,1] = Ntimes I$, where we are identifying $N$ with $Ntimes 0subseteq M$..



Now suppose $f:Mrightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:Ntimes Irightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:Mtimes Irightarrow X$ satisfying two conditions. First, for any $nin N$, $tin I$, that $G(n,t) = F(n,t)$. Second, for any $min M$, $G(m,1) = f(m)$.



To that end, set $G(m,t) = begincases f(m) & mnotin Ntimes [0,1]\ F(n, maxs,t) & m=(n,s)in Ntimes [0,1].endcases$



To check the first condition, note that $nin Ncong Ntimes 0$ means the corresonding $s$ is $s= 0$. Since $tin[0,1]$, $maxs,t = max0,t= t$, so $G(n,t) = F(n,t).$



To check the second condition, we have $G(m,t) = f(m)$ if $mnotin Ntimes [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)in Ntimes [0,1]$, then because we are taking $t=1$, we have $maxs,t = t$. So $G(m,1) = F(m,1) = f(m)$.



Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.



But, if $min Ntimes [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $maxs,t = s=1$, so $G(m) = F(n, maxs,t) = F(n,1) = f(n).$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    +1 for pointing out that you just need to verify the homotopy extension property.
    $endgroup$
    – William
    Mar 11 at 17:35










  • $begingroup$
    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
    $endgroup$
    – Moishe Kohan
    2 days ago










  • $begingroup$
    Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
    $endgroup$
    – Jason DeVito
    yesterday













8












8








8





$begingroup$

Yes, you get a deformation retract. More formally...




Suppose $M$ is a topological manifold, $Nsubseteq partial M subseteq M$ is a component of $partial M$ and that the inclusion $Nrightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.




Here's the idea of the proof:



From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.



Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.



So, some closed neighborhood of $N$ in $M$ is of the form $Ntimes [0,1] = Ntimes I$, where we are identifying $N$ with $Ntimes 0subseteq M$..



Now suppose $f:Mrightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:Ntimes Irightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:Mtimes Irightarrow X$ satisfying two conditions. First, for any $nin N$, $tin I$, that $G(n,t) = F(n,t)$. Second, for any $min M$, $G(m,1) = f(m)$.



To that end, set $G(m,t) = begincases f(m) & mnotin Ntimes [0,1]\ F(n, maxs,t) & m=(n,s)in Ntimes [0,1].endcases$



To check the first condition, note that $nin Ncong Ntimes 0$ means the corresonding $s$ is $s= 0$. Since $tin[0,1]$, $maxs,t = max0,t= t$, so $G(n,t) = F(n,t).$



To check the second condition, we have $G(m,t) = f(m)$ if $mnotin Ntimes [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)in Ntimes [0,1]$, then because we are taking $t=1$, we have $maxs,t = t$. So $G(m,1) = F(m,1) = f(m)$.



Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.



But, if $min Ntimes [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $maxs,t = s=1$, so $G(m) = F(n, maxs,t) = F(n,1) = f(n).$






share|cite|improve this answer











$endgroup$



Yes, you get a deformation retract. More formally...




Suppose $M$ is a topological manifold, $Nsubseteq partial M subseteq M$ is a component of $partial M$ and that the inclusion $Nrightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.




Here's the idea of the proof:



From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.



Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.



So, some closed neighborhood of $N$ in $M$ is of the form $Ntimes [0,1] = Ntimes I$, where we are identifying $N$ with $Ntimes 0subseteq M$..



Now suppose $f:Mrightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:Ntimes Irightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:Mtimes Irightarrow X$ satisfying two conditions. First, for any $nin N$, $tin I$, that $G(n,t) = F(n,t)$. Second, for any $min M$, $G(m,1) = f(m)$.



To that end, set $G(m,t) = begincases f(m) & mnotin Ntimes [0,1]\ F(n, maxs,t) & m=(n,s)in Ntimes [0,1].endcases$



To check the first condition, note that $nin Ncong Ntimes 0$ means the corresonding $s$ is $s= 0$. Since $tin[0,1]$, $maxs,t = max0,t= t$, so $G(n,t) = F(n,t).$



To check the second condition, we have $G(m,t) = f(m)$ if $mnotin Ntimes [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)in Ntimes [0,1]$, then because we are taking $t=1$, we have $maxs,t = t$. So $G(m,1) = F(m,1) = f(m)$.



Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.



But, if $min Ntimes [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $maxs,t = s=1$, so $G(m) = F(n, maxs,t) = F(n,1) = f(n).$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered Mar 11 at 17:29









Jason DeVitoJason DeVito

31.2k475136




31.2k475136











  • $begingroup$
    +1 for pointing out that you just need to verify the homotopy extension property.
    $endgroup$
    – William
    Mar 11 at 17:35










  • $begingroup$
    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
    $endgroup$
    – Moishe Kohan
    2 days ago










  • $begingroup$
    Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
    $endgroup$
    – Jason DeVito
    yesterday
















  • $begingroup$
    +1 for pointing out that you just need to verify the homotopy extension property.
    $endgroup$
    – William
    Mar 11 at 17:35










  • $begingroup$
    For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
    $endgroup$
    – Moishe Kohan
    2 days ago










  • $begingroup$
    Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
    $endgroup$
    – Jason DeVito
    yesterday















$begingroup$
+1 for pointing out that you just need to verify the homotopy extension property.
$endgroup$
– William
Mar 11 at 17:35




$begingroup$
+1 for pointing out that you just need to verify the homotopy extension property.
$endgroup$
– William
Mar 11 at 17:35












$begingroup$
For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
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– Moishe Kohan
2 days ago




$begingroup$
For the existence of collar neighborhoods of the boundary, an easier proof is also in Hatcher: proposition 3.42, p. 253. He proves this for compact manifolds but observes that the same proof works for paracompact manifolds.
$endgroup$
– Moishe Kohan
2 days ago












$begingroup$
Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
$endgroup$
– Jason DeVito
yesterday




$begingroup$
Moishe: Thanks for the reference! I had originally written this answer with $M$ a smooth manifold and realized the argument could generalize if topological manifolds with boundary had collars. I googled it, and that Annals paper came up. But it's nice to have a textbook reference!
$endgroup$
– Jason DeVito
yesterday

















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