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Theorem 2.13. If $X$ is a space and $A$ is a nonempty closed subspace that is a deformation retract of some neighborhood in $X$, then there is an exact sequence $$\cdots \longrightarrow \tilde H_n (A)\longrightarrow \tilde H_n(X) \longrightarrow \tilde H_n(X/A)\longrightarrow \tilde H_{n-1}(A)\longrightarrow \cdots $$

What's bugging me is the requirement $A$ be closed - isn't it enough to ask $\overline A\subset \mathring V$?

I ask the same question for the theorem stated in this homework sheet:

Theorem 0.2. Let $A\subset X$ be a closed subset such that $A$ is a deformation retract of some open set $V\subset X$. Then there is an isomorphism $$H(X,A)\cong H(X/A,\text{pt})$$

Isn't it enough to ask $\overline A\subset \mathring V$?

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If you look at the proof of proposition 2.22 in Hatcher's book, you see that there's a sequence of isomorphisms, some of which are valid by excision, and one of which is the map $H_n(X-A,V-A)\longrightarrow H_n(X/A-A/A,V/A-A/A)$ induced by the restriction of the quotient map $q:X\to X/A$ to the subspace $X-A$. You are right that excision works as long as we have $\overline A\subseteq \mathring V$. However, this restriction of $q$ to $X-A$ need not be a homeomorphism, but it certainly is when $A$ is closed (or open).

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  • $\begingroup$ Thank you, you pinpointed the issue. To show $q|_{A^c}\rightarrow (X/A)\setminus (A/A)$ given by $b\mapsto [b]={b}$ is a homeomorphism I must show it is bijective and bicontinuous. I (falsely of course) concluded this function is a homeomorphism regardless of whether $A$ is closed or open. I see why it is bijective, but as a corestriction of a continuous map, I think I'm not sure why it is continuous. Is this the part which requires $A$ to be open or closed (as a sufficient condition)? $\endgroup$ – user153312 Mar 3 '15 at 17:49
  • $\begingroup$ @Exterior: A restriction of a continuous map is always continuous. And if $A$ is closed, then $q|_{X-A}$ is an open map because when $U\subseteq X-A$ is open in $X-A$, then $U$ is open and saturated in $X$, so $q(U)$ is open. In case, $A$ is open, you can use closed sets of $X-A$. $\endgroup$ – Stefan Hamcke Mar 3 '15 at 17:59
  • $\begingroup$ Thanks for your reply. Can you elaborate a little on what 'saturated' means and how it's relevant? $\endgroup$ – user153312 Mar 3 '15 at 19:04
  • $\begingroup$ @Exterior: 'Saturated' means that $B=q^{-1}(q(B))$. In a quotient space a set is open if and only if it's the image of an open saturated set (the same holds with open replaced by closed). $\endgroup$ – Stefan Hamcke Mar 3 '15 at 19:30
  • $\begingroup$ Thanks for the clarification. In the first answer to this question, the author seems to say it suffices for $\overline A$ to have a neighborhood that deformation retracts to $A$ - without necessarily being closed - to have $H_n(X,A)=\tilde{H}_n(X/A)$. What am I missing here? $\endgroup$ – user153312 Mar 3 '15 at 20:07

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