Setup and conventions: Let $C_*$ be a chain complex of $R$-modules over some ring $R$, with boundary map $d$. The chain complex is said to be split if there exist $R$-linear maps $s: C_*\rightarrow C_{*+1}$ such that $d=dsd$. A chain map $f:C_*\rightarrow D_*$ is nullhomotopic if there exist $R$-linear maps $s:C_*\rightarrow D_{*+1}$ such that $f = ds + sd$; two chain maps are homotopic if their difference is nullhomotopic; and two complexes $C_*$ and $D_*$ are homotopy equivalent if there exist chain maps $f:C_*\rightarrow D_*$ and $g:D_*\rightarrow C_*$ such that $gf$ and $fg$ are homotopic to the identity maps respectively on $C_*$ and $D_*$. The homology $H_*(C)$ of a chain complex $C_*$ is itself a chain complex with all boundary maps zero, so that if $Z_*$ is the subcomplex of cycles and $B_*$ is the subcomplex of boundaries, $0\rightarrow B_*\rightarrow Z_*\rightarrow H_*(C)\rightarrow 0$ is an exact sequence in the category of chain complexes.
My question: Exercise 1.4.4 in Charles Weibel's Introduction to Homological Algebra asks us to provide an example of a chain complex that is homotopy equivalent to its homology, but is not split. After quite some thought about this, I have landed on what seems to me very much like a proof that it is impossible. But it's not plausible to me that Weibel either made a mistake or asked a trick question, so something must be wrong with my proof.
Can you help me find what's wrong with the proof?
My (must be faulty) proof: Let $C_*$ be the posited chain complex, and let $H_*(C)$ be its homology. By assumption we have chain maps $f:C_*\rightarrow H_*(C)$ and $g:H_*(C)\rightarrow C_*$ such that $gf$ and $fg$ are homotopic to the identity maps.
First consider $fg:H_*(C)\rightarrow H_*(C)$. We assume an $s':H_*(C)\rightarrow H_{*+1}(C)$ such that $1-fg = ds'+s'd$. However, $d:H_*(C)\rightarrow H_{*-1}(C)$ is zero, so $fg = 1$. Thus $f$ is surjective and $g$ is injective.
For each $n$, let $H_n'$ be the image of $H_n(C)$ in $C_n$ under $g$. Then $gf$ is a projection $C_n\rightarrow H_n'$: it is a projection because $(gf)^2=g(fg)f=gf$, and its image is $H_n'$ because $f$ is surjective.
Let $\pi = 1 – gf$. By assumption, $\pi$ is nullhomotopic (since $gf$ is homotopic to the identity), thus there exists $R$-linear $s:C_*\rightarrow C_{*+1}$ such that $\pi = ds + sd$. Since $gf$ is a projection, $\pi$ is a projection to $gf$'s kernel (which let's call $A_n$), and $C_n = A_n \oplus H_n'$.
Now consider the diagram
$$\begin{array} HH_n(C) & \xrightarrow{g} & C_n\\ \downarrow_{d=0} & & \downarrow_d\\ H_{n-1}(C) & \xrightarrow{g} & C_{n-1} \end{array}$$
which commutes by the assumption that $g$ is a chain map. Since $gd = g\circ 0 = 0$, $dg=0$ which means that $d$'s restricton to $\operatorname{im}g = H_n'$ is zero.
But this implies that if $c\in C_n$, then, with $c = a+h$ with $a = \pi(c)\in A_n$, $h=gf(c)\in H_n'$, we have $d(c) = d(a)$. And this is the same as saying that $d = d\pi$.
And $d\pi = d(ds + sd) = d^2s + dsd = dsd$ since $d^2=0$. Thus $d=dsd$, and $s$ realizes $C_*$ as split.
This looks like QED to me — what am I missing?
Note: this question is similar to this previous one, but it seemed worth asking both because that question is unanswered (update 8/23: I have answered it), and because I am asking a different question: I am specifically hoping for engagement with my (what must be faulty) logic, and I do not need an actual example of a nonsplit complex h.e. to its homology.
Addendum 8/22: I realized my argument can be given even more cleanly, without relying on the direct sum decomposition of $C_n$ or even needing to argue that $gf$ is a projection:
Assuming that $C_*$ is homotopy equivalent to its homology, let $f,g$ be as above. Then $dg=0$ simply because $g$ is a chain map, by the commutative diagram above. And $1-gf=ds+sd$ for some $R$-linear $s:C_*\rightarrow C_{*+1}$ because $f,g$ realize a homotopy equivalence. Then:
$$d = d-0 = d(1-gf) = d(ds+sd) = 0+dsd = dsd$$
because $d^2=0$. QED right?
Also, peter a g's comments seem to settle the question in favor of h.e. to homology $\Rightarrow$ split.
Best Answer
The 'must be faulty' proof (or proofs, with the addendum) above looks (look) correct, and the original exercise, as it appeared in the text, was wrong.
In fact - see http://www.math.rutgers.edu/~weibel/Hbook-corrections.html for corrections to Charles Weibel's Introduction to Homological Algebra. The page (currently) has links to the 1994 hard- and 1995 paper-back editions, and the paper-back link contains (direct quote)