I am trying to work through the paper Homological algebra of homotopy algebras by Hinich, but I don't know how to prove a certain claim he makes is true.
Let $k$ be a ring, $C(k)$ the category of complexes over $k$. Let $\mathcal{C}$ be a category, assume we have an adjoint pair of functors $$(-)^\#:\mathcal{C}\longleftrightarrow C(k):F,$$ and suppose that
- The catogory $\mathcal{C}$ has finite limits and colimits, and $\#$ commutes with filtered colimits.
- Let $d\in\mathbb{Z}$, define a complex $M\in C(k)$ by$$M^i:=\cases{k & $i = d,d+1$,\\0 & else,}$$ with differential given by the identity map between the two copies of $k$, and zero everywhere else. Then the canonical map $A\to A\sqcup F(M)$ induces a quasi isomorphism $A^\#\to(A\sqcup F(M))^\#$.
We define three classes of morphisms in $\mathcal{C}$:
- $W:=\{f\mid f^\#\text{ is a quasi isomorphism}\}$,
- $F:=\{f\mid f^\#\text{ is (degreewise) surjective}\}$,
- $C:=$ the set of maps with the left lifting property with respect to $F\cap W$.
We have the following construction: Let $A\in\mathcal{C}$, $M\in C(k)$, and $\alpha:M\to A^\#$ a map in $C(k)$. Denote by $cone(\alpha)$ the cone of $\alpha$ (i.e. the complex $M[1]\oplus A^\#$ with an appropriate differential) and by $\iota:A^\#\to cone(\alpha)$ the obvious inclusion. Take the pushout $$\require{AMScd} \begin{CD} F(A^\#) @>F(\iota)>> F(cone(\alpha))\\ @VVV @VVV\\ A @>>> A\langle M,\alpha\rangle \end{CD}$$ (first point of confusion: what is the left vertical map? I would expect the adjunction map going the other way round). In fact, this object represents a certain functor, denoted by $h_{A,\alpha}$ in the paper. But I don't think this is so relevant here.
Then at page 6 in the paper, Hinich claims that, if $M\in C(k)$ is a complex of free modules with zero differential, $A\in\mathcal{C}$ and $\alpha:M\to A$ are arbitrary, then the map $$A\longrightarrow A\langle M,\alpha\rangle$$ is a cofibration, that is to say in the class $C$ of maps defined above, and if instead $M$ is an acyclic complex of free modules, then the map is an acyclic cofibration, i.e. in $C\cap W$.
I tried a bit, applying $(-)^\#$ to the map and checking if the lifting property against the appropriate class holds. My idea was to use the fact that the functor preserves filtered colimits and the pushout universal property. However, I was not able to conclude (yet).
Any help, hint or solution would be greatly appreciated.
After reflection, I think that my first point of confusion is due to a typo in the paper. Philosophically, the category $\mathcal{C}$ is the category of some type of algebras over an operad (say dg commutative algebras, for example), $(-)^\#$ is the forgetful functor giving the underlying complex, and $F$ is the free algebra functor, which is left adjoint. Thus the left vertical map in the square is simply the counit of the adjunction.
My thoughts so far: let $X\to Y$ be a fibration (resp. an acyclic fibration) in $\mathcal{C}$. We consider the diagram: $$\require{AMScd} \begin{CD} F(A^\#) @>\epsilon_A>> A @>>> X\\ @VF(\iota)VV @VVV @VVV\\ F(cone(\alpha)) @>>> A\langle M,\alpha\rangle @>>> Y \end{CD}$$ Here $\epsilon$ denotes the counit of the adjunction. Now we apply $(-)^\#$ and add a small piece to the right: $$\require{AMScd} \begin{CD} A^\# @>\eta_{A^\#}>> F(A^\#)^\# @>(\epsilon_A)^\#>> A^\# @>>> X^\#\\ @V{\iota}VV @VF(\iota)^\#VV @VVV @VVV\\ cone(\alpha) @>\eta_{cone(\alpha)}>> F(cone(\alpha))^\# @>>> A\langle M,\alpha\rangle^\# @>>> Y^\# \end{CD}$$ with $\eta$ the unit of the adjunction. The middle square is still a pushout by condition (1) above, and the composition of the first two maps in the upper row gives identity. The rightmost map is surjective (resp. surjective and a quasi-isomorphism) by definition, so we can construct a diagonal filler $cone(\alpha)\to X$ (here I would like to draw e.g. a red diagonal arrow in the diagram, but I don't know how to do it with AMScd). Now I would really like to push it to the right a couple of times (but I would need the leftmost square to be a pushout to do it the first time), and then use it to induce a filler $A\langle M,\alpha\rangle\to X$ for the original diagram somehow.
I am unsure how to proceed. Maybe tomorrow I'll try to apply $F$ to the whole thing (after all, it is a left adjoint and so preserves colimits) to move my filler to the right once (using that the unit/counit composition gives identity), and then maybe using the counit for $A\langle M,\alpha\rangle$.
Ok, I think I almost got it. Apply $F$ to the last diagram above, and add once more a square on the left: $$\require{AMScd} \begin{CD} F(F(A^\#)^\#) @>\epsilon_{F(A^\#)}>> F(A^\#) @>F(\eta_{A^\#})>> F(F(A^\#)^\#) @>F((\epsilon_A)^\#)>> F(A^\#) @>>> F(X^\#)\\ @VVV @V{F(\iota)}VV @VF(F(\iota)^\#)VV @VVV @VVV\\ F(F(cone(\alpha))^\#) @>\epsilon_{F(cone(\alpha))}>> F(cone(\alpha)) @>F(\eta_{cone(\alpha)})>> F(F(cone(\alpha))^\#) @>>> F(A\langle M,\alpha\rangle^\#) @>>> F(Y^\#) \end{CD}$$ Now the first two maps in each of the rows compose to the identity, so we can precompose $F($our filler$)$ with $\epsilon_{F(cone(\alpha))}$ and then push it forward along the identity to get a map $F(F(cone(\alpha))^\#)\to F(X^\#)$, and then, using the fact that pushouts are preserved by $F$, push it forward once again to a map $F(A\langle M,\alpha\rangle)^\#)\to F(X^\#)$. Thus, we have obtained a filler for $$\require{AMScd} \begin{CD} F(A^\#) @>>> F(X^\#)\\ @VVV @VVV\\ F(A\langle M,\alpha\rangle^\#) @>>> F(Y^\#) \end{CD}$$ Now apply the counits on the right to get a filler for $$\require{AMScd} \begin{CD} F(A^\#) @>>> X\\ @VVV @VVV\\ F(A\langle M,\alpha\rangle^\#) @>>> Y \end{CD}$$ Now we only need a way to take away the $F((-)^\#)$ on the left.
Ok, let's finish this. Apply once more $(-)^\#$ to the last diagram and add to the left: $$\require{AMScd} \begin{CD} A^\# @>\eta_{A^\#}>> F(A^\#)^\# @>>> X^\#\\ @VVV @VVV @VVV\\ A\langle M,\alpha\rangle^\# @>\eta_{A\langle M,\alpha\rangle^\#}>> F(A\langle M,\alpha\rangle^\#)^\# @>>> Y^\# \end{CD}$$ This gives a map $A\langle M,\alpha\rangle^\#\to X^\#$. But $$\hom(A\langle M,\alpha\rangle^\#,X^\#)\cong\hom(A\langle M,\alpha\rangle,F(X^\#))$$ by adjunction, so we have in fact a map $$A\langle M,\alpha\rangle\longrightarrow F(X^\#).$$ Now postcompose with the unit $\epsilon_X$ of the adjunction. This gives a map with the correct domain and range, and it should fill the square as we wanted.
Now, this is probably much, much more complicated than it could. I applied the functors something like 6 times... Therefore, my question now becomes:
Is there an easier/shorter/more elegant way to do this?