Contractibility of an exact chain complex How can one prove that an exact (acyclic) chain complex of projective modules that is trivial in negative degrees is contractible?
I would appreciate some nudges in the right direction more than anything; I'm new to homological algebra and am having a hard time seeing how to start proving this.
 A: The hypothesis that the chain complex is bounded below points you in the direction of induction. Let $C_*$ be the chain complex. The chain complex being acyclic means that every cycle is a boundary, i.e. if $x \in C_n$ is such that $dx = 0$, then $x = dy$ for some $y \in C_{n+1}$. What we want is $C_*$ being contractible, meaning that there is a chain homotopy $h_n : C_n \to C_{n+1}$ such that $x = h_{n-1}(d_n x) + d_{n+1}(h_n x)$ for all $x \in C_*$.
It's this $h$ that's constructed by induction. The first one is $h_0 : C_0 \to C_1$ and it must satisfy $d_1 h_0 x = x$ for all $x \in C_0$ (because $C_{-1} = 0$). What are our hypotheses? The complex $C_*$ is acyclic, and $C_{-1} = 0$. So every $x \in C_0$ is a cycle ($C_{-1} = 0$), hence a boundary $C_*$ acyclic. Thus $d_1 : C_1 \to C_0$ is surjective.
And now we use the hypothesis that all the $C_n$ are projective modules. $d_1 : C_1 \to C_0$ is surjective and $C_0$ is projective, hence there is a lift of the identity, a map $s : C_0 \to C_1$ such that $d_1 s x = \operatorname{id}_{C_0}(x) = x$. This map $s$ is exactly the $h_0$ we needed.
And now we start again and want to construct $h_1 : C_1 \to C_2$ such that $x = h_0(d_1(x)) + d_2(h_1(x))$ for all $x \in C_1$. Recalling that we'll use the fact that the modules are projective, we rewrite the equation as
$$d_2(h_1(x)) = x - h_0(d_1(x)) = g(x),$$
where $g(x) = x - h_0(d_1(x))$.
To use projectivity, we need a surjective map. Let $Z_1 \subset C_1$ be the kernel of $d_1$. Then since $C_*$ is a chain complex, $d_2(C_2) \subset Z_1$; the acyclicity of $C_*$ means that $d_2 : C_2 \to Z_1$ is surjective.
The map we want to lift is $g : C_1 \to C_1$, so all we need to check is that $g(C_1) \subset Z_1$ and we're done. But using $d_1(h_0(y)) = y$, we find:
$$d_1(g(x)) = d_1(x) - d_1(h_0(d_1(x)) = d_1(x) - d_1(x) = 0$$
and so $g(C_1) \subset Z_1$. We can now use projectivity of $C_1$ to find a lift $h_1 : C_1 \to C_2$ such that $d_2(h_1(x)) = g(x) = x - h_0(d_1(x))$.

There are a few details missing here, and you need to write down the induction fully. I'll let you fill in the details.
