Let $C\_{\ast} : \cdots \to A\_{2} \to A\_{1} \to A\_{0} \to 0$ be a chain complex of finite dimensional vector spaces over a field $K$.


And let $f\_{\ast} : C\_{\ast} \to C\_{\ast}$ and $g\_{\ast} : C\_{\ast} \to C\_{\ast}$ be two chain endomorphisms of $C\_{\ast}$ satisfying $\det (f\_{n}) \neq 0$ and $\det (g\_{n}) \neq 0$ for any $n$.


Assume that the homology group $H\_{n}(C\_{\ast})$ is zero for all but finitely many $n$.
Then, the induced homomorphism $H\_{n}(f\_{\ast}) : H\_{n}(C\_{\ast}) \to H\_{n}(C\_{\ast})$ is zero for all but finitely many $n$, and so the alternating product $\prod\_{n}\det (H\_{n}(f\_{\ast}))^{(-1)^{n}}$ is well-defined (of course, the same holds for $g\_{\ast}$).


Moreover, assume that $\det (f\_{n}) = \det (g\_{n})$ for any $n$.


My question is:




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QUESTION


Under the above conditions, does the next equation hold up to sign?


$\prod\_{n} \det (H\_{n}(f\_{\ast}))^{(-1)^{n}} = \prod\_{n} \det (H\_{n}(g\_{\ast}))^{(-1)^{n}}$.




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Note that if the chain complex $C\_{\ast}$ is bounded above, the statement can be proved as follows:


$\prod\_{n} \det (H\_{n}(f\_{\ast}))^{(-1)^{n}} = \prod\_{n} \det (f\_{n})^{(-1)^{n}} = \prod\_{n} \det (g\_{n})^{(-1)^{n}} = \prod\_{n} \det (H\_{n}(g\_{\ast}))^{(-1)^{n}}$


Here, the first and the last equation can be shown by using induction on the length of $C\_{\ast}$, snake lemma and the multiplicativity of $\det$ for short exact sequences. The middle equation is the result of the assumption.


The problem is that, in general, $\prod\_{n} \det (f\_{n})^{(-1)^{n}}$ and $\prod\_{n} \det (g\_{n})^{(-1)^{n}}$ are not well-defined.


By Sawin's answer
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There is a counterexample if one does not admit the difference of sign.


Please give me any advice.



Endomomorphisms of Chain Complexes of vector spaces and determinants
This is false. Let $C\_\*$ be the complex where each $C\_i$ is two-dimensional and all the maps have rank one, so that there is homology only in degree $0$. Let $f$ be the identity, and let $g$ be multiplication by $-1$. Then since they are acting on two-dimensional vector spaces, their determinants are the same everywhere. But the homology group is one-dimensional, so the determinants of their action on homology are $+1$ and $-1$ respectively, which are different.


