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In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where

  • weak equivalences are the quasi-isomorphisms,
  • fibrations are degree-wise surjections, and
  • cofibrations are defined by the left lifting property.

Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.

Question: Is it true that in this case every chain complex is also cofibrant?

I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?

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Yes.

Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form $$\dots\to0\to k\to0\to\dots$$ and $$\dots\to0\to k\stackrel{\sim}{\to}k\to0\to\dots$$ and their shifts.

It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.

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it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.

The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.

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