Let $\dots \to 0 \to M_n\to M_{n-1}\to\dots\to M_0\to 0 \to \dots$ be a complex of projective modules. Prove that this complex is acyclic (zero homology) iff the complex is isomorphic to the direct sum of complexes of the form $M\xrightarrow{id}M$ with shifts.
(Note: The shifted complex $M[r]$ is the complex which in degree n is $M_{n+r}$, with differential $d_{M[r]} = (−1)^rd_M$.)
I have proved that a finite complex of projective is acyclic iff the identity morphism is homotopic to the 0 morphism, but I'm not sure if this is on the right track.
Look at the final map $\pi:M_1\to M_0$. This is a surjection, so splits, because $M_0$ is projective. There's a splitting map $i:M_0\to M_1$ with $\pi\circ i =\text{id}_{M_0}$. Then $M_1$ is the direct sum of $\ker\pi=M_1'$ and $\text{im}\, i=M_0'$. Both these are projective, and the original exact sequence splits into two exact sequences $$0\to M_n\to M_{n-1}\to\cdots\to M_2\to M_1'\to0\to0$$ and $$0\to 0\to 0\to\cdots\to 0\to M_0'\to M_0\to0.$$ Now induct on the length of sequence.