Are all (unbounded) chain complexes over a field bifibrant?

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?

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