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Question: Let $\mathbb{F}$ be an algebraically closed field of characteristic zero and let $\mathrm{Ch}_{\mathbb{F}}$ be the category whose objects are chain complexes (of $\mathbb{F}$-modules) and whose morphisms are chain maps. Is it true that every object in $\mathrm{Ch}_{\mathbb{F}}$ is injective? If so, does it remain true if $\mathbb{F}$ is not algebraically closed? What if $\mathbb{F}$ has positive characteristic?

On one hand this seems to good to be true, but on the other hand things are often exceedingly nice when dealing with vector spaces.

Edit: removed a flawed proof. Thanks darij for spotting the error.

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  • $\begingroup$ "the above sequence gives rise to sequences of $\mathbb F$-modules in each degree:" -- how so? A homomorphism of chain complexes is not just a collection of homomorphisms between their components; it has to send every component to the respective component of the image, and commute with the boundary operator. $\endgroup$
    – darij grinberg
    Commented Feb 24, 2015 at 19:38
  • $\begingroup$ @darji As you say a chain complex morphism is a collection of F-module maps which commute with the boundary operator. The fact that they commute with the boundary operator isn't really relevant at this point, all that matter is that they are linear maps. So if I have a chain map f_{.}:A_{.}->B_{.}, then I have linear maps f_{n}:A_{n}->B_{n}. By definition of image and kernel in the above proof, such maps should compose to give an exact sequence in each degree. Perhaps I'm misunderstanding your concern? $\endgroup$
    – Paul
    Commented Feb 24, 2015 at 19:53
  • $\begingroup$ What do you mean by "such maps should compose to give an exact sequence in each degree"? From $f_\cdot$ you don't get an exact sequence. How do you obtain $0\to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(C_{m},K_{n}\big) \to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(B_{m},K_{n}\big) \to \mathrm{Hom}_{\mathrm{Ch}_{\mathbb{F}}}\big(A_{m},K_{n}\big) \to \mathrm{D}_{m,n}$ ? $\endgroup$
    – darij grinberg
    Commented Feb 24, 2015 at 19:55
  • $\begingroup$ @darji I see now, you're correct; this part of the proof looks wrong, I'll have to think about it some more. I'm less interested in the correctness of the proof than in the answer to the question. Is there an object in Ch_{F} which is not injective? $\endgroup$
    – Paul
    Commented Feb 24, 2015 at 20:14

2 Answers 2

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No.

Let $X$ be the chain complex with $\mathbb{k}$ in degree $n$ (and all differentials zero) and let $Y$ be the chain complex with $\mathbb{k}$ in degrees $n$ and $n+1$ with the identity as differential $Y_{n+1}\to Y_n$ (I choose for my convention that the differential lowers degree).

Then $X$ is not injective because there is no chain map $Y\to X$ which factors the identity on $X$ through the evident inclusion of $X$ into $Y$.

To be injective over a field it is necessary and sufficient to have vanishing homology.

A similar argument shows that not every chain complex is projective, and that to be projective over a field it is necessary and sufficient to have vanishing homology.

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  • $\begingroup$ Exactly what I was looking for. Thanks! $\endgroup$
    – Paul
    Commented Feb 24, 2015 at 20:48
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    $\begingroup$ In fact, every chain complex over a field is isomorphic to a direct sum of shifts of these two complexes $X$ and $Y$. $\endgroup$
    – Eric Wofsey
    Commented Feb 24, 2015 at 21:02
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More abstractly, in an abelian category, saying either that every object is injective or that every object is projective is equivalent to saying that every short exact sequence splits (semisimplicity).

Categories of chain complexes essentially never have this property, e.g. because the chain complex $c \xrightarrow{{id}_c} c$ is always a nontrivial extension of its degree-$0$ and degree-$1$ parts provided that $c$ is not itself zero.

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