Let $f$ be a morphism of chain complexes. Show that if $ker(f)$ and $coker(f)$ are acyclic, then $f$ is a quasi-isomorphism.

Question

Let $f$ be a morphism of chain complexes. Show that if $ker(f)$ and $coker(f)$ are acyclic, then $f$ is a quasi-isomorphism. Is the converse true?

I am self reader of homology algebra and I stuck in this,it will be great if you help me,thanks.

this is exercise 1.3.5 of Weibel’s book “An Introduction to Homological Algebra”,by the way.

I will explain my problem here:

consider this sequence :

$$...\rightarrow ker f_{n+1} \overset{b_{n+1} |_{kerf_{n+1}}}{\rightarrow} ker f_{n} \overset{b_n |_{kerf_{n}}}{\rightarrow} ker f_{n-1} \rightarrow...$$

which $f:(C_{.},b_{.}) \rightarrow (D_{.},e_{.})$

and

$$...\rightarrow coker f_{n+1} \overset{\widetilde{e}_{n+1}}{\rightarrow} coker f_{n} \overset{\widetilde{e}_{n}}{\rightarrow} ker f_{n-1} \rightarrow...$$

which

$$\widetilde{e}_{n}:\frac{D_{n}}{Imf_{n}} \rightarrow \frac{D_{n-1}}{Imf_{n-1}} $$

$$x+Imf_{n} \rightarrow e_{n}(x)+ Imf_{n-1}$$

I must show that $$H_{n}(f) :H_{n}(C_{.}) \rightarrow H_{n}(D_{.})$$

$$x+Im e_{n+1} \rightarrow f_{n}(x)+Im d_{n+1}$$

is isomorphism,it will be great if you help me about it,thanks.

Answer 1

We have the exact sequences

$$(1) \quad 0 \to \ker f \to C \to \mathrm{im }f \to 0$$

and

$$(2) \quad 0 \to \mathrm{im }f \to D \to \mathrm{coker }f \to 0.$$

The long exact sequence of homology coming from $(1)$, along with the assumption that $\ker f$ is acyclic, shows that $C \to \mathrm{im} f$ is a quasi-isomorphism.

The long exact sequence of homology coming from $(2)$, along with the assumption that $\mathrm{coker} f$ is acyclic, shows that the inclusion $\mathrm{im} f \to D$ is a quasi-isomorphism.

Hence the composite $C \to \mathrm{im} f \to D$, which is $f$, is a quasi-isomorphism.