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Let $A,B$ be differential modules with differentiation homomorphism $d$ (such that $d^2=0$). Then let say that $g$ is an epimorphism from $A$ into $B$. Then is it possible for an induced homomorphism $H(g):H(A)\to H(B)$ to not be an epimorphism? ($H(C)=Z(C)/B(C)$)

Right now, I think it is possible, but I don't see the clear way to show it. I think there's no reason for $a+B(A)$ to be in $H(A)$ for arbitrary $b+B(C)\in H(C)$, where and $g(a)=b$.

Any kind of advice and help are welcome!

generic properties
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1 Answers1

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I use the convention "epi=surjective". By definition, the map $g:A\rightarrow B$ is a surjective morphism of degree $0$ of differential complexes of modules; then $g\circ d_A=d_B\circ g$. The statement about surjectivity is true if

$$g^{-1}(Z(B))\subseteq Z(A), (*) $$

in each degree. In fact, at the level of cohomology, each element of $H^i(B)$ is of the form $b+d_B\tilde{b}$ with $b\in B^i$ s.t. $d_B(b)=0$, for some $\tilde{b}\in B^{i-1}$. So

$$b+d_B\tilde{b}=(\text{surjectivity})=g(a)+d_B(g(\tilde{a}))=(\text{def. of morphism of complexes})=g(a+d_A(\tilde{a}))=H(g)([a]),$$

for some $a\in Z^i(A)\subseteq A^i$, $\tilde{a}\in A^{i-1}$ if $(*)$ is true.

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  • Although the notions of the derived modules of higher degree are not familiar to me (I'm studying with Sze-Tsen Hu's book), I get from your answer that I have to negate $()$ in order to show that my hypothesis is true. Now, for general cases, usually $()$ does not hold, I think not assuming surjectivity is plausible. (Am I right?) But also, w/o additional info. of A,B I cannot assume no surjectivity too, I guess. Thank you. – generic properties Dec 04 '13 at 14:54