0
$\begingroup$

Let $\mathcal{A}$ be an abelian category with enough injectives. If $X^{\bullet}$ is a bounded below complex, it is a well known fact that you can obtain a bounded below complex $I^{\bullet}$ of injective objects in $\mathcal{A}$ and a quasi-isomorphism $\epsilon: X^{\bullet} \rightarrow I^{\bullet}$. The answer to the question here gives a particularly nice construction of this, but I am not quite sure I follow the logic.

The construction proceeds inductively. It shows that if you have the complex $X^{\bullet}$, and some complex of injectives $I^{\bullet}$ with morphisms $\{ f^{n}: X^{n} \rightarrow I^{n} \text{ for all } n \leq a \}$, then you can inductively produce $f^{n+1}$ so that it is a quasi-isomorphism in each degree.

This is all well and good, but I don't see how you pass from the finite to the infinite. This allows you to construct a sequence of truncated above chain complexes, with quasi-isomorphisms in higher and higher degrees, but unless you have some notion of "taking the limit" then I don't see how this produces a morphism of complexes. The obvious notion here would be some kind of direct limit, but this construction is supposed to work even when direct limits may not be exact.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.