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Let $M$ be an abelian group, and consider a “projective resolution” $$ 0 \to M \to P_0 \to P_1 \to \cdots \to P_n \to 0. $$ Note that this is not a projective resolution in the usual sense since it's going the wrong way! But I claim that there must exist a projective resolution $P_\bullet \to M$ going the right way. The argument is as follows. The exactness of the above sequence exhibits a quasi-isomorphism $M \to P_\bullet$. This arrow must be invertible in the derived category, and since $P_\bullet$ is “cofibrant” (here I am using the boundedness of $P_\bullet$), there must exist a map $P_\bullet \to M$ realizing this inverse. Thus, $P_\bullet$ is a projective resolution of $M$ in the usual sense.

Is this correct? I'm mainly asking as a sanity check since this situation seems so strange to me.

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  • $\begingroup$ I doubt when you say that $P_{\bullet}$ is a projective resolution in the usual sense. Since the map $P_{n-1} \to P_n$ is already surjective and $P_0 \to P_1$ is not injective, $P_{\bullet}$ is not a resolution. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ I don't follow the argument. Sure, there's a map, but the arrows in $P_\bullet$ do not magically change directions: You get a map from $P_0$ to $M$, and a map from $P_0$ to $P_1$, and a map from $P_1$ to $P_2$, etc. You seem to be suggesting that there should be a map $P_0 \to M$, and a map $P_1 \to P_0$, and a map $P_2 \to P_1$, but where are these wrong-way maps supposed to come from? $\endgroup$ Commented 9 hours ago

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What you claim is correct, but not particularly interesting, since the only $M$ that have a "projective resolution" in the sense you describe are the projective ones (the map $P_{n-1}\to P_n$ splits, and now work by induction), so the "projective resolution" is of the form $$0\to M\to M\oplus Q_0\to Q_0\oplus Q_1\to\dots\to Q_{n-1}\oplus Q_n\to Q_n\to 0.$$

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