EDIT After thinking carefully with the help of the clear answer of ZhenLin, I think I will reformulate my question the following way. The text of my original question is kept below.

The claim of Prof. Rotman in section 10.5 that split complexes (not necessarily exact) with projective terms are the projectives in the category of chain complex relies on the result of exercise 10.15.

In this exercise, it is asked to prove that if $$0\to A'\to A\to A''\to 0$$ is split exact with $\delta$ the injection $A'\to A$, then the complex $$0\to A'\to A\to 0$$ concentrated in degrees $k$ and $k-1$ with the only non zero differential $\delta$ is a direct summand of the complex $$0\to A\to A\to 0$$ concentrated in degrees $k$ and $k-1$ with the only non zero differential $1_A$.

This exercise looks innocuous but I cannot solve it and I believe it is wrong.

If I am not mistaken, this explain why the approach of Prof. Rotman is wrong, that is to say the projectives in the chain complex category must be split exact complexes, not only split ones. Therefore Corollary 10.37 & Theorem 10.42 are wrong, and also the fact that Cartan-Eisenberg resolutions are projectives ones!

But perhaps I am confused. Does somebody can give me his informed opinion here?

Old question

In Prof. Weibel AIHA book exercise 2.2.1 it is asked to show that the projectives of the chain complex category of a category of modules are the split exact complexes of projectives.

In Prof. Rotman AIHA book theorem 10.42 it is proved that the projectives of the chain complex category of an abelian category are the split complexes of projective, not necessarily exact! There are some points in his proof that I do not understand:

1/ He uses the Freyd-Mitchell embedding on an infinite numerable direct sum (Prop. 10.36). Is it legitimate ? Edit: there should be a workaround here because locally in each degree of the complex the infinite direct sum is finite.

2/ In his proof of Corollary 10.37, he said that direct sum of projectives are projectives. Is it true in any abelian category ? Edit2: yes, clearly as Zhen Lin pointed out in his answer.

If we work in a module category, I can understand his proof. So who is right ? Edit3: there must be something wrong in the proof even in a category of modules, since the answer of ZhenLin is crystal clear. I have to look in details at it.

  • $\begingroup$ Any abelian category (with some "smallness" requirement) can be identified with a subcategory of a category of modules. $\endgroup$ – Tobias Kildetoft Feb 9 '16 at 19:35
  • $\begingroup$ @TobiasKildetoft Thank you for your precision. But does it mean that every infinite direct sums stays in this subcategory ? $\endgroup$ – brunoh Feb 9 '16 at 19:38
  • $\begingroup$ @TobiasKildetoft and my main question is about who is right since I cannot detect mistakes in the proof of Prof. Rotman in a module category ... $\endgroup$ – brunoh Feb 9 '16 at 19:40
  • $\begingroup$ Well, since the category of modules is closed under arbitrary direct sums, yes. $\endgroup$ – Tobias Kildetoft Feb 9 '16 at 19:40
  • 1
    $\begingroup$ You're right, the statement that is to be proved in Exercise 10.15 is false. The complex $0\to A\to A\to 0$ is exact, so any direct summand must be exact. But the complex $0\to A'\stackrel{\delta}{\to} A\to0$ is only exact if $\delta$ is an isomorphism. $\endgroup$ – Jeremy Rickard Feb 11 '16 at 9:33

The statement for which a proof is asked in Exercise 10.15 is false.

The complex $$\dots\to0\to A\stackrel{1_A}{\to}A\to0\to\dots$$ is exact, and so any direct summand is also exact. But unless $\delta$ is an isomorphism, the complex $$\dots\to0\to A'\stackrel{\delta}{\to}A\to0\to\dots$$ is not exact, and so is not a direct summand of the first complex.

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Let $\mathcal{A}$ be an abelian category and let $\mathcal{C}$ be the category of unbounded chain complexes in $\mathcal{A}$. For each integer $n$, we have a functor $(-)_n : \mathcal{C} \to \mathcal{A}$ sending a chain complex $C_\bullet$ to the object $C_n$, and it has both a right adjoint, namely the functor sending an object $A$ to the chain complex $\cdots \to 0 \to A \to A \to 0 \to \cdots$ concentrated in degrees $n+1$ and $n$. The right adjoint preserves epimorphisms, so the left adjoint preserves projectives. In particular, projective objects in $\mathcal{C}$ are degreewise projective.

It remains to be shown that projective objects in $\mathcal{C}$ are exact. (A degreewise-projective exact chain complex is automatically split.) Suppose $P_\bullet$ is a projective object in $\mathcal{C}$. Following the hint given in Weibel, we have an exact complex $Q_\bullet$ and an epimorphism $Q_\bullet \to P_\bullet$; but every epimorphism into a projective object splits, so $P_\bullet$ is a summand of $Q_\bullet$, hence $H_* (P) = H_* (Q) = 0$, as desired.

I leave it to you to show that split-exact degreewise-projective chain complexes are projective objects in $\mathcal{C}$. The fact that coproducts of projective objects are projective is true in any category whatsoever and will be useful for this.

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  • $\begingroup$ Perfectly crafted and clear answer as usual. I am so grateful to you for taking the time. Just for my sake, I will add in the last sentence "in any cocomplete abelian category". So to summarize, Prof. Rotman seems to be wrong and Prof. Weibel is right. I just have to find where precisely! $\endgroup$ – brunoh Feb 9 '16 at 21:31
  • $\begingroup$ If I understood well the proof of Prof. Rotman, the gist of it is to say that if $P$ and $Q$ are projectives, the complex $$0\to P\to P\oplus Q\to 0$$ where the differential is $1_P$ is projective in the category of chain complex. It is clearly a non-exact split complex so why is it projective ? $\endgroup$ – brunoh Feb 9 '16 at 23:40
  • $\begingroup$ No, there is no need to add anything to my last assertion. There is a notion of projective object in any category, and coproducts of projective objects – whenever they exist – are automatically projective. $\endgroup$ – Zhen Lin Feb 10 '16 at 0:21
  • $\begingroup$ You are obviously right. I don't know what I was thinking. $\endgroup$ – brunoh Feb 10 '16 at 0:32

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