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I am having trouble understanding a corollary in Lang's algebra. In particular, corollary 3.2 on page 770. I have been given the definition for the Euler-Poincaré characteristic of a complex $E^{\bullet}$ with respect to $\phi$ to be $$ \chi_{\phi}(E^{\bullet}) = \sum(-1)^{i}\phi(H^{i}) .$$ where $H^{i}$ is the homology of the complex. A theorem then states that if $F^{\bullet}$ is a complex (which is of even length if it is closed) and we assume that $\phi(F^{i})$ is defined for all $i$ and is equal to $0$ for all but finitely many of those $i$, and that $H^{i}(F^{\bullet}) = 0$ for all but finitely many $i$, then we have $$ \chi_{\phi}(F^{\bullet}) = \sum(-1)^i \phi(F^{i}). $$ Finally, as a corollary of this theorem, Lang states that if $F^{\bullet}$ is an acyclic complex, such that $\phi(F^{i})$ is defined for all $i$ and equal to $0$ for all for all but finitely many $i$, and that if $F^{\bullet}$ is closed then it is of even length, then we have that $$ \chi_{\phi}(F^{\bullet}) = 0. $$

My question is, why does he even bother with the first theorem? Isn't it trivially $0$ since the complex is assumed to be acyclic? And why does it have to be of even length for the corollary? By the definition, if all the homology groups are trivial (which is what being acyclic means), then straight away the alternating sum of them is $0$. What am I missing?

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    Maybe it's worth recalling what is $\phi$ here. In Lang's terminology, it's an "Euler-Poincaré mapping", which means that (1) for each short exact sequence $0 \to M' \to M \to M'' \to 0$ the value $\phi (M)$ is defined if and only if $\phi (M')$ and $\phi (M'')$ are both defined, and $\phi (M) = \phi (M') + \phi (M'')$ and (2) $\phi (0) = 0$. – 2017-01-07
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    As by definition $\phi (0) = 0$, and for Lang $\chi_\phi (F^\bullet)$ is defined in terms of cohomology, then for acyclic $F^\bullet$ we indeed automatically get $\chi_\phi (F^\bullet) = 0$. I guess Lang was trying to say something different — for instance, define first $\chi_\phi (F^\bullet)$ as $\sum (-1)^i\,\phi (F^i)$, then show that $\sum (-1)^i\,\phi (F^i) = \sum (-1)^i\,\phi (H^i)$, and then state the corollary for acyclic $F^\bullet$. It's just my guess. – 2017-01-07

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