# If a chain complex is homotopy equivalent to its homology, is it split?

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.

• The only place I can see where the logic doesn't seem airtight is the sentence "since $gf$ is a projection, $\pi$ is a projection to $gf$'s kernel,... and $C_n = A_n \oplus H'_n$". I'm not convinced that $\pi$ surjects onto $A_n$, and thus there may be some weird problem preventing the direct sum decomposition from being true. I'd try to look for counterexamples by using sequences of abelian groups which you know don't split as direct sums, and work through that step in your argument carefully. – Dorebell Aug 22 '15 at 2:12
• So far, I have to say that your argument looks brilliant to me - is this p 18 line 3? See: math.umd.edu/~jmr/602/bookerrors.pdf – peter a g Aug 22 '15 at 2:46
• And here is a link to the author's site with more corrections math.rutgers.edu/~weibel/Hbook-corrections.html - the above (with p 18 line 3) is for the paper back version; this page has a link to the same pdf as above. – peter a g Aug 22 '15 at 3:13
• @Dorebell - I added an addendum that gives the argument without relying on this decomposition, but in any case I'm sure of it. $\pi$ acts as the identity on any element of $A_n$, so it surjects. General claim: if $A$ is a module (over any ring) and $\tau\in\operatorname{End}A$ is a projection (i.e. $\tau^2=\tau$), then $1-\tau$ is a projection to $\tau$'s kernel, and $A = \ker\tau \oplus \operatorname{im}\tau$. Proof: $(1-\tau)^2 = 1-2\tau + \tau = 1-\tau$, so $1-\tau$ is a projection; if $x\in\ker \tau$, then $(1-\tau)(x) = x$, so $1-\tau$ acts as the identity on $\ker \tau$; cont'd... – Ben Blum-Smith Aug 22 '15 at 13:20
• @Dorebell - and if $y\in\operatorname{im}(1-\tau)$ i.e. $y=(x-\tau x)$ then $\tau y = \tau x - \tau^2 x = 0$, so $1-\tau$ maps into $\ker\tau$. (Because it acts as the identity on it, it must be onto.) For the direct sum decomposition, $x = \tau(x) + (1-\tau)(x)$ gives any $x\in A$ as the sum of $\tau(x)\in \operatorname{im}\tau, (1-\tau)(x)\in\ker \tau$, and the representation is unique since $\ker\tau \cap \operatorname{im}\tau = 0$ since $\tau$ acts trivially on the former and as the identity on the latter. – Ben Blum-Smith Aug 22 '15 at 13:24