If $A$ is a weak deformation retraction of $X$, is $i*$ an homology isomorphism? Let $A \subseteq X$ be a weak deformation retraction as in this definition: https://topospaces.subwiki.org/wiki/Weak_deformation_retraction
Does this mean that $i* : H_n(A) \to H_n(X)$ is an isomorphism? I think I know how to prove this if $A$ is a deformation retraction, but not in this weaker sense.
 A: By definition of weak deformation retraction, there exists a continuous function $H:X\times I \to X$, where $I=[0,1]$, such that
(i) $H(x,0)=x$ for every $x$ in $X$.
(ii) $H(a,t)$ is in $A$ for every $(a,t)$ in $A\times I$.
(iii) $H(x,1)$ is in $A$ for every $x$ in $X$.
Now, consider the corestriction $f:x\in X \mapsto H(x,1) \in A$, which is well defined by (iii). We will prove that $f$ is an homotopy inverse of $i$ and so $i$ is an homotopy equivalence.
(1) Consider $f\circ i : A \to A$. The restriction and corestriction $F: (a,t)\in A\times I \mapsto H(a,t) \in A$ is well defined by (ii). Moreover, by (i) and the definition of $f$, $H$ is an homotopy between the identity map $id_A$ and $f\circ i$.
(2) Consider $i\circ f: X\to X$. Note that $H$ is an homotopy between the identity map $id_X$ and $i\circ f$.
Thus, by (1) and (2), $f$ is an homotopy inverse of $i$ and then, in particular, $i$ is an homotopy equivalence.
By a basic result from algebraic topology, $i$ induces an homotopy equivalence of chain complex $C_*(A)\to C_*(X)$ and then, by a basic result from homological algebra, $i$ induces an isomorphism $H_*(A)\to H_*(X)$ at the level of the homology. 
