3

This is the second half of exercise 1.3.5 in An Introduction to Homological Algebra by Weibel, it simply asks if this statement is true of false and I believe it is false but cannot construct a counterexample.

Ler $f:A \rightarrow B$ be a morphism of chain complexes.

The first half is proving the converse of this statement which goes by we have exact sequences $0 \rightarrow kerf \rightarrow A \rightarrow Imf \rightarrow 0$ and $0 \rightarrow imf \rightarrow B \rightarrow cokerf \rightarrow 0$. These give us two long exact sequences in Homology and because $kerf$ and $cokerf$ are acyclic we get isomorphisms $H_n(A) \cong H_n(imf) \cong H_n(B)$ for all $n$ so that we have a quasi isomorphism as desired.

Now for my question assume that $f$ is a quasi-isomorphism. We still have these exact sequences and thus the two long exact sequences in Homology. So we have $...\rightarrow H_n(kerf) \rightarrow H_n(A) \xrightarrow{i_*} H_n(imf) \rightarrow H_{n-1}(kerf) \rightarrow ...$ and $... \rightarrow H_{n+1}(cokerf) \rightarrow H_n(imf) \xrightarrow{j_*} H_n(B) \rightarrow H_n(cokerf) \rightarrow ...$.

Where $f_* = (j \circ i)_* = j_* \circ i_*$. Now for $f_*$ to be an isomorphism all this does is force my $j_*$ to be $1-1$ and $i_*$ to be surjective so that it forces my induced morphisms $H_n(kerf) \rightarrow H_n(A)$ and $H_n(B) \rightarrow H_n(cokerf)$ to be zero. Now I do not see why the induced morphisms having to be zero requires that my homology for the kernel and cokernel are 0. Any insights would be appreciated!

Michael N
  • 355

2 Answers2

7

Considering the following two short exact sequences

$$0\rightarrow 0 \rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$$ and $$0\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}\rightarrow 0\rightarrow 0$$ where the maps from $\mathbb{Z}$ to $\mathbb{Z}$ are identity and other maps are trivial.

Then we can construct a chain complex morphism $f$ between them: the only non-trivial map is the middle one which is identity.

Then $\ker f$ is $$0\rightarrow 0 \rightarrow 0\rightarrow \mathbb{Z}\rightarrow 0$$ and $\mathrm{coker}\ f$ is $$0\rightarrow \mathbb{Z} \rightarrow 0\rightarrow 0\rightarrow 0$$ and they are not acyclic. However $f$ is a quasi-isomorphism.

Yining Zhang
  • 717
5

Suppose $f:C\to D$ is a quasi-isomorphism. If $f$ is either onto or injective, then the remaining (co)kernel must be acyclic by a LES argument. Thus for you to find a counterexample, you need to find a quasi-isomorphism that is neither injective nor a surjection to begin with.

Consider the complex $D$ where $D_n= \Bbb Z_4$ and $\partial$ is multiplication by $2$. Consider now the complex $C$ that has $\Bbb Z_2\oplus\Bbb Z_2$ in each degree, and the diferential are the projections in each coordinate, alternately. Both this complexes are acyclic. Consider now the map of complexes $f:C\to D$ that is the inclusion of $\Bbb Z_2$ in $ \Bbb Z_4$ in alternating coordinates, so that $f$ is a morphism. Then the kernel of $f$ is a the complex with $\Bbb Z_2$ in each degree and zero differential, so it is not acyclic. The cokernel of $f$ is also $\Bbb Z_2$ in each degree, with zero differential. Thus $f$ is trivially a quasi isomorphism (because $0\simeq 0$), but it has nonacyclic kernel and cokernel.

Pedro
  • 122,002