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I'm having an issue with exercise 2.4.2 in Weibel's homological algebra book. Namely, it asks you to prove that if $U:\cal{B}\rightarrow\cal{C}$ is an exact functor, then

$$U(L_iF)\simeq L_i(UF)$$

Where $L_iF$ is the left derived functor of $F:\cal{A}\rightarrow\cal{B}$. I can prove equivalence of homology, but assuming he means isomorphism in terms of natural transformations I'm struggling to prove the commutativity of the associated diagram for a map $f: A\rightarrow A'$ in $\cal{A}$. I guess my issue is while it's easy to prove the existence of the isomorphism between homology groups, getting an explicit form is less so, so trying to follow elements through the diagram is awkward. Thanks in advance for your help!

ottoak
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1 Answers1

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Suppose $F:A\to B$ is the functor you want to derive, and $U:B\to C$ is exact. To compute $LF(a)$ for ann object $a$ of $A$, you consider a projective resolution $a_*\stackrel{\varepsilon}\longrightarrow a $, and then compute $HF(a_*)$. Then you obtaon $ULF(a)$ simply by applying $U$ to the result, $UHF(a_*)$. To compute $LUF(a)$, you want to consider $HUF(a_*)$.

It follows that what you really want to prove is the following: if $b_*$ is a complex in $B$ then there is an isomorphism $H(Ub_*)\to UH(b_*)$. To do this, it suffices you show that $U$ commutes with taking homology $H=Z/B$ at one spot. But we have an exact sequence $0\to B\to Z\to H\to 0$, and applying $U$ you get an exact sequence $0\to UB\to UZ\to UH\to 0$. This identifies $HU$ with $UH$ canonically.

Pedro
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  • Right, I was able to get that far. That an exact functor commutes with homology on objects makes sense, but my issue is showing the commutativity of the diagram involving a map $f:A\rightarrow A'$ in $\cal{A}$ that a natural transformation must satisfy. – ottoak Sep 27 '18 at 00:53
  • @ottoak The identification induced by the exact sequence is natural, since a morphism $f:A\to A'$ induces a morphism on resolutions and then a morphism of the "fundamental" short exact sequences BZH. This is all natural up to the choice of projective resolutions of $A$ and $A'$. – Pedro Sep 28 '18 at 00:15