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Need surjective chain map induce surjective homomorphism between homology groups? Since $f_*[x]\to[f(x)]$?

Daniel Xu
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No, consider the morphism of chain complexes of vector spaces over a field $k$ $$C:= ... \to 0\to k \to k \to 0\to ...$$ and $$C':= ... \to 0 \to k \to 0 \to 0\to ...$$ where the first $k$ is in degree $0$ for both complexes, and consider $f: C \to C'$ as the obvious non zero morphism between those two ($id$ in degree $0$ and $0$ otherwise). Then this defines a surjective morphism of chain complexes, but since $H^*(C)=0$ and $H^0(C')=k$ we get that $H^*(f)$ can't be surjective.

The problem with your reasoning is that your preimage might die in homology, i.e not even be represented.

Felix
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  • vvery thankful! – Daniel Xu Nov 07 '18 at 14:16
  • Is this example really correct? What is the operator $\partial:k\to k$?
    1. If it is $Id$ the identity, then $f\partial =Id$ while $\partial f = 0$ and then $f$ is not a chain map.
    2. If it is $0$, then $H^0(C)$ is also $k$, instead of $0$.
     – Derso Aug 29 '20 at 12:19
  • It is the identity, and $f$ is going from $C$ to $C'$ so both $f \circ d =0 =d \circ f$ just for degree reasons. I am quite sure you confused the direction of the arrow. Actually what is happening here is that no matter what you do, a square ending in $0$ always commutes. – Felix Aug 30 '20 at 19:38