1
$\begingroup$

Let's suppose that $F\colon \mathcal A\to \mathcal B$ is a right exact additive functor between abelian categories such that $\mathcal A$ has enough projectives. Standard references shows that if $Q_\bullet\to A\to 0$ is a resolution of $A$ by $F$-acyclic objects, then $L_nF(A)\cong H_n(F(Q_\bullet))$. Suppose now that $f\colon A\to A'$ is a morphism in $\mathcal A$ and $Q'_\bullet \to A'\to 0$ is a resolution of $A'$ by $F$-acyclic objects. Suppose moreover, I have a chain map $Q_\bullet\to Q'_\bullet$ over $f$. Then I get induced homomorphism $f_n\colon H_n(F(Q_\bullet))\to H_n(F(Q'_\bullet))$ which agree with $f$ at the level of $H_0$ (after the natural identifications). My question is can we choose isomorphisms $L_nF(A)\to H_n(F(Q_\bullet))$ and $L_nF(A')\to H_n(F(Q'_\bullet))$ so that these isomorphisms intertwine $f_n$ and $L_nF(f)$ for all $n$?

Any pointers to references would be appreciated.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Looks like you forgot to apply $F$ to your resolutions before taking homology. $\endgroup$
    – user509184
    Commented Mar 21 at 22:23
  • 1
    $\begingroup$ @user509184, Sorry I will fix $\endgroup$
    – Benjamin Steinberg
    Commented Mar 21 at 23:04
  • $\begingroup$ Here is how I faintly recall this working, but I haven't tried to write anything down to be careful about it, so caveat emptor! If your resolutions $Q$ and $Q'$ were projective resolutions, then it's standard that the map induced by $f$ in $L_*F$ is the one induced by $H_*(f)$ from $A$ to $A'$. So you just need to compare $Q$ and $Q'$ to projective resolutions. I think the standard approach is to take Cartan-Eilenberg resolutions of $Q$ and $Q'$, and use projectivity of the CE resolutions to give you a map between them which is induced by $f$ and yet also induces the right map in $L_*F$. $\endgroup$
    – user509184
    Commented Mar 21 at 23:20
  • $\begingroup$ @user509184 I was trying something along that line but my problem was to check that that the map from the projective resolution to Q is a quasi-isomorphism. The proof I’m used to that F-acyclic resolutions work is usually based on dimension shifts. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 22 at 0:28
  • 1
    $\begingroup$ I think that it suffices to assume that $\mathcal A$ has enough $F$-acyclic objects (e.g. sheaf categories usually do not have enough projectives). $\endgroup$
    – Z. M
    Commented Mar 22 at 11:29

0

Reset to default

You must log in to answer this question.