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

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    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.2012-02-01
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    Apparently not -- see Weibel Introduction to Homological Algebra p43.2012-02-01
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    Hm, that's the book I learned from! Let me look...2012-02-01
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    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 $H_0(F(P)) = F(P_0)/\operatorname{im}(F(P_1) \to F(P_0)) \approx F(A)$.2012-02-01
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    Sorry, how did you get that last isomorphism?2012-02-01
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    Just from the last displayed exact sequence that you wrote down: $F(P_0)$ surjects onto $F(A)$ and that's the kernel.2012-02-01
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    Ah right, because the sequence is exact, $\mathrm{im}(F(P_1)\to F(P_0))=\mathrm{ker}(F(P_0)\to F(A))$.2012-02-01
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    @DylanMoreland Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2014-06-03

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