# Are two homotopic maps between chain complexes equal up to composition with a homotopy equivalence?

Let $f,g : A\rightrightarrows B$ be two maps between chain complexes $A,B$ with terms in an abelian category. Suppose $f,g$ are homotopic. Must there exist a homotopy equivalence $\alpha : C\rightarrow A$ such that $f\circ \alpha = g\circ \alpha$ on the nose? (also, the same question with $\beta : B\rightarrow C$ such that $\beta\circ f = \beta\circ g$)

• What is $C$? Does it have any relations to $A,B$? There seems to be some missing information here. – KReiser Dec 13 '16 at 0:38
• There is no relation. $C$ is just another chain complex. – user355183 Dec 13 '16 at 16:07

No. For instance, in the category of vector spaces over a field $$k$$, let $$A$$ be the chain complex $$0\to k\to 0$$ (with $$k$$ in degree $$0$$) and let $$B$$ be the chain complex $$0\to k\to k\to 0$$ (with the $$k$$'s in degrees $$1$$ and $$0$$). Let $$f$$ be the inclusion map $$A\to B$$ and let $$g$$ be the zero map $$A\to B$$. Then $$f$$ and $$g$$ are homotopic, since $$B$$ is contractible. But if $$\alpha:C\to A$$ is any homotopy equivalence, then $$\alpha$$ must be surjective (in order to be surjective on homology), and so $$f\alpha$$ and $$g\alpha$$ cannot be equal.
For homotopy equivalences $$B\to C$$, you can use the same example but in the opposite category. (Or, the quotient map and the zero map $$B\to B/A$$ give a counterexample for the category of vector spaces, since any homotopy equivalence $$B/A\to C$$ must be injective.)