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Let $R$ be a commutative ring with unity. Let $A_i$ be an R-module for every $i$. Consider a sequence of modules $$\xrightarrow{\delta_{i-1}}A_{i-1}\xrightarrow{\delta_{i}} A_i\xrightarrow{\delta_{i+1}} A_{i+1}.$$

Let a morphism of a sequence of modules be defined as in this question, which does not use exactness. I am wondering whether an isomorphism of sequences preserves exactness at a module $A_i$ in the sequence.

I cannot find an counterexample, but exactness is a set-theoretical property (kernel of $\delta_i$ is equal to the image of $\delta_{i-1}$ as subsets of the module $A_{i-1}$) which does not have to be preserved by an isomorphism.

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up vote 2 down vote accepted

As a rule of thumb, isomorphisms preserve everything. So why would you think exactness might not be preserved? In fact, you will get induced isomorphisms for the kernels and cokernels (the universal properties defining them are somewhat "blind" against isomorphisms). This doesn't even require a concrete category (i.e. argumenting with set elements).

More generally, if the sequence is a not necessarily exact complex (i.e. $\delta_i\circ \delta_{i-1}=0$ for all $i$), an siomorphism of complexes induces an isomorphism of homologies (i.e. of the quotients $\ker \delta_i/\operatorname{im}\delta_{i-1}$).

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