Why is $0 \to H_n(A) \to H_n(X) \to H_n(X/A) \to 0$ exact

algebraic-topologyexact-sequencehomology-cohomology

Let $X$ be a topological space and $A$ a subspace of $X$. Is

$$0 \to H_n(A) \to H_n(X) \to H_n(X/A) \to 0$$

exact at the middle homology group? Here $i_*: H_n(A) \to H_n(X)$ is the map induced by the inclusion $i: A \to X$ and $q_*:H_n(X) \to H_n(X/A)$ is the map induced by the quotient map $q: X \to X/A$.

I was able to show that $\operatorname{Im}(i_*) \subseteq \operatorname{Ker}(q_*)$ but I wonder if the other inclusion also holds?

If not, can we say more if $A$ is a deformation retract of an open subset of $X$?

Best Answer

Yes this is true but only if $i$ is a cofibration. Use the long exact sequence for the pair $(X,A)$ and the identity $H_n(X,A)=H^r_n(X/A)$ where the $r$ stands for reduced homology.

Beware though, that identity generally only holds if $i$ is a cofibration.