Taking zeroth cohomology of a dg-module over a dg-category is "good" on representables, but probably "bad" on cones

Question

This question is possibly related to this other one.

Let $\mathcal A$ be a dg-category over a commutative ring $k$. I denote by $\text{dgm-}\mathcal A$ the dg-category of right dg-$\mathcal A$-modules, that is, dg-functors \begin{equation} \mathcal A^{\text{op}} \to \mathbf C_{\text{dg}}(k), \end{equation} where $\mathbf C_{\text{dg}}(k)$ is the dg-category of cochain complexes of $k$-modules. In general, the category of dg-functors $\mathrm{Fun}_{\text{dg}}(\mathcal A, \mathcal B)$ between two dg-categories is itself a dg-category, with hom-complexes denoted by $\mathrm{Nat}_{\text{dg}}(F,G)$ (the complex of natural transformations $F \to G$).

Let me recall the (differential graded) Yoneda lemma. Given $F \in \text{dgm-}\mathcal A$, we have a natural isomorphism of complexes: \begin{equation} \mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F) \cong F(A). \end{equation} We also have the differential graded Yoneda embedding: \begin{align} h_{-} : \mathcal A \hookrightarrow \text{dgm-}\mathcal A, \\ A \mapsto \mathcal A(-,A). \end{align}

A useful operation on dg-modules is taking cohomology. In fact, there is a functor \begin{equation} H^0(-) : H^0(\text{dgm-}\mathcal A) \to \text{mod-}H^0(\mathcal A), \end{equation} which takes a dg-module $F$ to the $H^0(\mathcal A)$-module of its zeroth cohomology.

Now, notice that the zeroth cohomology of a representable module $\mathcal A(-,A)$ is given by $H^0(\mathcal A(-,A)) = H^0(\mathcal A)(-,A)$. So, this $H^0(\mathcal A)$-module is also represented by $A$. Then, we can apply both the differential graded and the ordinary version of Yoneda lemma: \begin{align} & H^0(\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F)) \cong H^0(F(A)) \\ &= H^0(F)(A) \cong \mathrm{Nat}(H^0(\mathcal A(-,A)), H^0(F)), \end{align} and it is immediate to see that this chain of maps gives actually the functorial map \begin{equation} H^0(\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F)) \to \mathrm{Nat}(H^0(\mathcal A(-,A)), H^0(F)) \end{equation} given by the above $H^0(-)$ functor. So, we have proven that in this case, this natural map is an isomorphism.

Now, let me take another step. I start from a closed degree zero map $f: A \to B$ in $\mathcal A$, which corresponds to a closed degree zero morphism $h_f : h_A \to h_B$ via the Yoneda embedding. The category $\text{dgm-}\mathcal A$ is (strongly) pretriangulated, so we can take the cone $C(h_f)$, which fits in the following pretriangle: \begin{equation} h_A \xrightarrow{h_f} h_B \to C(h_f) \to h_A[1]. \end{equation} Now, let $F \in \text{dgm-}\mathcal A$, and consider the dg-functor \begin{equation} \mathrm{Nat}_{\text{dg}}(-, F) : (\text{dgm-}\mathcal A)^{\text{op}} \to \mathbf{C}_{\text{dg}}(k). \end{equation} As a dg-functor, it maps pretriangles to pretriangles (caveat: because of contravariance, the sign of the shift is changed to the opposite). Hence, we obtain a pretriangle in $\mathbf{C}_{\text{dg}}(k)$: \begin{equation} \mathrm{Nat}_{\text{dg}}(h_B,F) \to \mathrm{Nat}_{\text{dg}}(h_A,F) \to \mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \to \mathrm{Nat}_{\text{dg}}(h_B[-1],F). \end{equation} Moreover, Yoneda lemma gives a commutative square:

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so we obtain a (functorial and strict!) isomorphism between cones: \begin{equation} \mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \cong C(F(f)) \end{equation} as cochain complexes. Rearranging shifts, we obtain: \begin{equation} \mathrm{Nat}_{\text{dg}}(C(h_f),F) \cong C(F(f))[-1]. \end{equation} Now, I'd like to compare $H^0(\mathrm{Nat}_{\text{dg}}(C(h_f),F)) \cong H^0(C(F(f))[-1])$ with $\mathrm{Nat}(H^0(C(h_f)),H^0(F))$. Is it true or false that the natural map \begin{equation*} H^0(\mathrm{Nat}_{\text{dg}}(C(h_f),F)) \to \mathrm{Nat}(H^0(C(h_f)),H^0(F)) \end{equation*} is an isomorphism? I expect the answer to be no, because, roughly speaking, taking cohomology "destroys" the functoriality of cones. How can I find a counterexample?

Answer 1

You can take your DG-category to be a plain ring $R$, or the category of f.g. projective $R$-modules. Then you're asking the following question: Given a complex of f.g. projective $R$-modules $C$ concentrated in degrees $-1$ and $0$ and an arbitrary complex $F$, is $H^0(\hom(C,F))\cong \hom(H^0(C),H^0(F))$? Actually you're asking whether a specific map is an isomorphism, but I'll show you an example where there is no even an abstract isomorphism. Assume that the projective dimension of $R$ is $>0$ (otherwise your map is indeed an iso). Take two $R$-modules $M$ and $N$, $M$ finitely presented, such that $\operatorname{Ext}_R^1(M,N)\neq 0$. Let $C$ be a presentation of $M$ and $F=N[-1]$, in particular $H^0(F)=0$. However $H^0(\hom(C,F))$ surjects onto $\operatorname{Ext}_R^1(M,N)$, so it cannot be trivial.