Let $F:\mathcal{A}\to\mathcal{B}$ be a covariant right-exact functor between two abelian categories.

Suppose $\mathcal{A}$ has enough projectives. Then we define the left derived functors of $F$ by $$ L_iF(A)=H_i(F(P_\bullet)) $$ where $A$ is any object in $\mathcal{A}$ and $P_\bullet$ is a projective resolution for $A$ (it can be shown that $L_iF(A)$ is independent of the choice of projective resolution.

Since $F$ is right-exact, the sequence $$ F(P_1)\to F(P_0)\to F(A)\to 0 $$ is exact. Doesn't this mean that $L_0F(A)=H_0(F(P_\bullet)))=0$, since the homology of an exact complex is zero? However everywhere I look says that $L_0F(A)\cong F(A)$.

  • 4
    $\begingroup$ One usually "chops off" the $A$ and takes the homology of $\cdots \to F(P_1) \to F(P_0) \to 0$, right? I thought that $L_0F(A) = F(A)$ was more or less an axiom. $\endgroup$ Feb 1, 2012 at 2:01
  • $\begingroup$ Apparently not -- see Weibel Introduction to Homological Algebra p43. $\endgroup$ Feb 1, 2012 at 3:17
  • $\begingroup$ Hm, that's the book I learned from! Let me look... $\endgroup$ Feb 1, 2012 at 3:18
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    $\begingroup$ I think that Weibel and I agree, although he's not being super clear on this point. You're taking $H_i(F(P))$; the augmented complex with $P_0 \to A \to 0$ on the end is another thing. He writes that right exact sequence just to show that $H_0(F(P)) = F(P_0)/\operatorname{im}(F(P_1) \to F(P_0)) \approx F(A)$. $\endgroup$ Feb 1, 2012 at 3:24
  • 2
    $\begingroup$ Just from the last displayed exact sequence that you wrote down: $F(P_0)$ surjects onto $F(A)$ and that's the kernel. $\endgroup$ Feb 1, 2012 at 3:44

1 Answer 1


I think this is easier to understand if you rephrase your definition of a projective resolution.

Let $A_{\bullet}$ be a chain complex. Then, a projective resolution of $A_{\bullet}$ is a quasiisomorphism (that is, a morphism of chain complices that induces an isomorphism on homology) $$ \newcommand{\arr}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\ard}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllllll} \cdots & \arr{} & P_2 & \arr{} & P_1 & \arr{} & P_0 \\ & & \ard{} & & \ard{} & & \ard{} \\ \cdots & \arr{} & A_2 & \arr{} & A_1 & \arr{} & A_0 \end{array} $$ with each $P_m$ projective. (Taking $A_1=A_2=\cdots =0$ gives you something equivalent to the 'classical' definition.) The idea is that these two chain complices are isomorphic in the derived category (the localization of the category of chain complices at the quasiisomorphisms), and so to compute the derived functor applied to $A_{\bullet}$, we can apply it to $P_{\bullet}$ instead. It turns out that $F(P)_{\bullet}$ is quasiisomorphic to $LF(A)_{\bullet}$ (whereas $F(A)_{\bullet}$ is not in general), and so if all your interested in is the homology of the (total) left derived functor $LF(A)_{\bullet}$, $F(P)_{\bullet}$ will work just fine. Anyways, back to the original question.

In the case your interested in, applying $F$ to the above diagram yields $$ \begin{array}{lllllll} \cdots & \arr{} & F(P_2) & \arr{} & F(P_1) & \arr{} & F(P_0) \\ & & \ard{} & & \ard{} & & \ard{} \\ \cdots & \arr{} & 0 & \arr{} & 0 & \arr{} & F(A_0) \end{array} $$ The classical left derived functors are the homology of the top chain complex. Drawing things this way makes it obvious where the indexing should start (certainly not at $F(A_0)$). It also makes it conceptually clearer why one would perform such a construction.

Taking homology of this diagram gives in particular a map $LF_0(A):=H_0(F(P))\rightarrow F(A_0)$. It then turns out that this defines a natural transformation $LF_0\rightarrow F$ which is a natural isomorphism iff $F$ is right-exact. In particular, that $F(P_0)\rightarrow A_0\rightarrow 0$ is exact if $F$ is right-exact isn't really relevant.


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