Chain map factoring through 0 map on Homology

Question

Suppose you have chain complexes $A,B,C,D$. And say you have maps $f:A \to B, g:B \to C, h:C \to D$, but only $g$ is a chain map. However, $h\circ g\circ f$ is a chain map even though $f$ and $h$ weren't chain maps.

Now suppose that $g$ induces the $0$ map on Homology. Then does it follow in some way that $h\circ g\circ f$ induces the $0$ map on Homology?

I imagine it should be easy to come up with a counterexample playing with chain complexes of small-dimensional vector spaces with only a few nonzero terms. — Eric Wofsey, Jan 25 '20 at 02:51
Did you understand my example? Let me know if anything is not clear. — Cheerful Parsnip, Jan 26 '20 at 15:59

score 4 · Accepted Answer · answered Jan 25 '20 at 19:49

Let $A\cong B\cong D$ be $1$ dimensional concentrated in a single degree, say $0$. Let $C$ be of the form $0\to C_2\to C_1\to 0$ where $C_2$ and $C_1$ are both $1$ dimensional and the boundary map is an isomorphism. Let $g,h,f$ just carry the generators in degree $1$ to each other. Then clearly the composition $h\circ g\circ f$ is a chain map and the induced map on homology is the identity. Moreover $g$ is also clearly a chain map. So this is a counterexample to the conjecture.

Chain map factoring through 0 map on Homology

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