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I've been trying to wrap my head around derived functors and have come upon the following chain of arguments, which seems to yield the unreasonabel conclusion that all left derived functors are exact. I presume there must be some mistake, but have thus far been unable to find it, and would be gratefull for any help. The argument goes like this. Take some exact sequence $$0 \rightarrow A \rightarrow X \rightarrow C \rightarrow 0 $$ With projective resolutions $$A_* \rightarrow A, C_*\rightarrow C$$ By the horseshoe lemma we may choose a projective resolution $X_*\rightarrow X$ st. $X_n = A_n \oplus C_n$, s.t the sequence of deleted deleted chain complex $$A_*\rightarrow X_* \rightarrow C_*$$ with connecting arrows respectively the injection and projection of the coproduct, is a split short exact sequence of chain complexes. We may compute our left derived functor on our original sequence using this sequence, and since derived functors are only defined on additive functors, homology functors are also additive, and additive functors preserve split exact sequences, this seems to suggest that the derived functor on any short exact sequence yields a split short exact sequence. A conclusion that seems mildly utopic, if any one can point out what i did wrong it would be much appreciated.

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The middle complex $X_\ast$ has terms that are $A_n\oplus C_n$, but it is generally not true that $X_\ast=A_\ast\oplus C_\ast$ as complexes. To compute the left derived functors of $F$ you would apply $F$ to each of the three resolutions and take the homology. It is indeed true that you will have split exact sequences $$0\to F(A_n)\to F(A_n)\oplus F(C_n)\to F(C_n)\to 0,$$ but the differential in the middle column will not be the sum of differentials of the outer columns. As a result, you will not get a split exact sequence $$0\to H_n(F(A_\ast))\to H_n(F(A_\ast))\oplus H_n(F(C_\ast))\to H_n(F(C_\ast))\to 0.$$

To have a look at what's going on, you can try to write the three resolutions explicitly for the exact sequence $$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0$$

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