Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $L_0F(A)\cong F(A)$.

share|cite|improve this question
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. – Dylan Moreland Feb 1 '12 at 2:01
Apparently not -- see Weibel Introduction to Homological Algebra p43. – Clinton Boys Feb 1 '12 at 3:17
Hm, that's the book I learned from! Let me look... – Dylan Moreland Feb 1 '12 at 3:18
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 $H_0(F(P)) = F(P_0)/\operatorname{im}(F(P_1) \to F(P_0)) \approx F(A)$. – Dylan Moreland Feb 1 '12 at 3:24
Just from the last displayed exact sequence that you wrote down: $F(P_0)$ surjects onto $F(A)$ and that's the kernel. – Dylan Moreland Feb 1 '12 at 3:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.