It is well known that cohomology takes short exact sequences to long exact sequences. Moreover, the connecting morphisms are natural meaning that cohomology takes morphisms of short exact sequences to those of long exact sequences. Now, let \begin{array}[llll]0 0 &\to &A &\to &B &\to &C &\to &0\\ &&\downarrow &&\downarrow &&\downarrow\\ 0 &\to &D &\to &E &\to &F &\to &0\\ \end{array} be a diagram of chain complexes where squares commute only up to chain homotopy. Does cohomology functor take this diagram to a morphism of long exact sequences? Thanks.
Naturality of connecting morphisms
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homological-algebra
1 Answers
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Let $0\to A\to B\to C\to0$ be any short exact sequence of chain complexes such that $B$ is contractible (chain homotopy equivalent to the zero complex), but $A$ and $C$ have non-zero homology.
For example, the short exact sequence $$0\to[\dots\to0\to\mathbb{Z}\to\dots]\to [\dots\to\mathbb{Z}\stackrel{\sim}{\to}\mathbb{Z}\to\dots]\to [\dots\to\mathbb{Z}\to0\to\dots]\to0$$ of complexes of abelian groups.
Then any choice of maps $A\to A$, $B\to B$, $C\to C$ gives a diagram that commutes up to chain homotopy. But if you choose the zero map $A\to A$ and the identity map $C\to C$, then this doesn't induce a morphism of long exact sequences.