Proving the isomorphism in homology

Question

Let $A$ be a chain complex and let $B \subseteq A$ be a sub complex. Assume there is a chain map $\alpha: A \to A$ such that the following conditions are fulfilled

(i) $\alpha$ is chain homotopic to the identity map on $A$

(ii) $\alpha(B) \subseteq B$, and $\alpha_{\big|B}:B \to B$ is chain homotopic to the identity on $B$.

(iii) For every $a \in A_n$, there is an $m$ such that $\alpha^m(a) \in B_n$

How can I prove that the inclusion $B \to A$ induces an isomorphism on homology?

I have tried to solve this exercise for some days, but with no further success, I posted a similar question before but didn't get any answers or hints. So far this is what I have done:

I have proved that the map $f: H_n(B) \to H_n(A)$ is a well defined map and I have proved that $\alpha^m$ will be homotopic to the identity as $\alpha$ is, however from here I have no idea how to prove the bijectivity, for the map being injective I have tried to prove that its kernel is trivial, i.e if $f(a)=0$ then $a=0$, this gives us a lot of information about $a$, for instance it is in the image and kernel of the differentials restricted to the $B_n$ but it is also in the image of $d_{n+1} : A_{n+1} \to A_{n}$, but I can't seem to prove that $a$ must be in the image of $d_{n+1}$ restricted to the submodule $B_n \subseteq A_n$.

EDIT: I have yet no proof for the injectivity but the surjectivity can be proven as follows: First, note that if given two homotopic chain maps $f_1 \simeq f_2: A_{\ast} \to B_{\ast}$ and $g_1 \simeq g_2: B_{\ast} \to C_{\ast}$ then $g_1 \circ f_1 \simeq g_2 \circ f_2: A_{\ast} \to C_{\ast}$. This allows us to conclude that since $\alpha \simeq \text{Id}$ we must have $\alpha^m \simeq \text{Id}$.

Homotopic chain maps induces the same maps on the homology groups of the chains, that is $$\alpha_{\ast}^m = \text{Id}_{\ast}: \text{ker}(d_n)/ \text{Im}(d_{n+1}) \to\text{ker}(d_n)/ \text{Im}(d_{n+1}).$$ Let $x \in \text{ker}(d_n)/ \text{Im}(d_{n+1})$, then $$x = \text{Id}_{\ast}(x)=\alpha_{\ast}^m(x),$$ that is $\alpha^m(x) = x$ in $\text{ker}(d_n)/ \text{Im}(d_{n+1})$ but $\alpha^m(x) \in B_n$ so in the map $H_n(B) \to H_n(A)$ induced by the inclusion we have $\alpha^m(x) \mapsto x$ and so surjectivity is obtained.

SECOND EDIT:

The injectivity can be proven as follows:

Consider the injection $$f: \text{ker}(d_{n}^{\ast})/ \text{Im}(d_{n+1}^{\ast}) \to \text{ker}(d_{n})/ \text{Im}(d_{n+1}),$$ suppose that $f(a)=0$, we have to prove that $a \in \text{Im}(d_{n+1}^{\ast})$. Since $\alpha$ and the identity is homotopic, then so is $\alpha^m$ and the identity, in particular the induce the same map in homology, i.e $$\text{Id}= \alpha^m : \text{ker}(d_{n}^{\ast})/ \text{Im}(d_{n+1}^{\ast}) \to \text{ker}(d_{n}^{\ast})/ \text{Im}(d_{n+1}^{\ast}),$$ so that proving that $a \in \text{Im}(d_{n+1}^{\ast})$ is equivalent to proving that $\alpha^m(a) \in \text{Im}(d_{n+1}^{\ast})$. It exists an $x \in A_{n+1}$ such that $d_{n+1}(x)=a$, thus by the properties of having a chain map we conclude that $$\alpha^m(a)=\alpha^m(d_{n+1}(x))=d_{n+1}(\alpha^m(x)),$$ however $\alpha^m(x) \in B_{n+1}$, we conclude that $\alpha^m(a) \in \text{Im}(d_{n+1}^{\ast})$ and so $a \in \text{Im}(d_{n+1}^{\ast})$ which completes the proof.

Answer 1

One can show that the inclusion $i:B\to A$ is an homotopy equivalence if $A$ is a chain complex of free modules and $B$ is another free subcomplex. Indeed, let $h$ be such that $1-\alpha=dh+hd$. For each basis element $a\in A$, pick the first $m=m_a$ such that $\alpha^m(a)\in B$. Now define $H$ on basis elements to be $h(1+\alpha+\cdots+\alpha^{m_a-1})(a)$, so that $H(a) = 0$ if and only if $a\in B$. Since $\alpha$ is a chain map, we have that $m_{\delta a} \leqslant m_a$ for each $a\in A$.

Now one observes that $D =1+dH+Hd$ has image in $B$ (this follows from a careful manipulation and the definition of $m_a$) that $Di = 1_B$ and that $iD$ is homotopic to the identity of $A$. You can check the details in Spanier's book on algebraic topology, page 178, theorem 14. This implies your result, because we can always do the following (when $A,B$ are chain complexes over a PID...): given $i:B\to A$ an inclusion of possible non-free complexes, we can construct new complexes $FB$ and $FA$ of free modules, an inclusion $Fi : FB\to FA$ and maps $FB\to B$ and $FA\to A$ such that the following commutes

$$ \require{AMScd} \begin{CD} 0 @>>> B @>{i}>> A\\ {} @A\varepsilon_BAA @A\varepsilon_{A }AA \\ 0 @>>> FB @> Fi>> FA \end{CD} $$ and the vertical maps are surjective quasi-isomorphisms. Since then $H(i)H(\varepsilon_B) = H(Fi)H(\varepsilon_A)$ and all maps but possibly $H(i)$ are isomorphisms, so is $H(i)$.

Answer 2

Let $\alpha_* : H_n (A) \to H_n (A)$ denote the induced homology map. Since $\alpha^k (a) \in B_n$ for sufficiently large $k \in \mathbb{N}$, we deduce that $\alpha^k (A_n ) \subset B_n$ for $k$ large enough.

In particular, we have that $\big( \alpha^k \big)_* : H_n (A) \to H_n (B)$. But, as $\alpha$ is chain homotopic to the identity, we know that $\alpha_* = \textrm{id}$. By (covariant) functoriality of the association $\alpha \mapsto \alpha_*$, $$\big( \alpha^k \big)_* = \big( \alpha_* \big)^k = \textrm{id}^k = \textrm{id}= \alpha_*$$ Whence $H_n (A) \cong H_n (B)$.