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

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 $$$$\mathcal A^{\text{op}} \to \mathbf C_{\text{dg}}(k),$$$$ where $$\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: $$$$\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F) \cong F(A).$$$$ 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 $$$$H^0(-) : H^0(\text{dgm-}\mathcal A) \to \text{mod-}H^0(\mathcal A),$$$$ which takes a dg-module $$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 $$$$H^0(\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F)) \to \mathrm{Nat}(H^0(\mathcal A(-,A)), H^0(F))$$$$ given by the above $$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: $$$$h_A \xrightarrow{h_f} h_B \to C(h_f) \to h_A[1].$$$$ Now, let $$F \in \text{dgm-}\mathcal A$$, and consider the dg-functor $$$$\mathrm{Nat}_{\text{dg}}(-, F) : (\text{dgm-}\mathcal A)^{\text{op}} \to \mathbf{C}_{\text{dg}}(k).$$$$ As a dg-functor, it maps pretriangles to pretriangles (caveat: because of contravariance, the sign of the shift is changed to the opposite). Hence, we obtain a pretriangle in $$\mathbf{C}_{\text{dg}}(k)$$: $$$$\mathrm{Nat}_{\text{dg}}(h_B,F) \to \mathrm{Nat}_{\text{dg}}(h_A,F) \to \mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \to \mathrm{Nat}_{\text{dg}}(h_B[-1],F).$$$$ Moreover, Yoneda lemma gives a commutative square:

(source: presheaf.com)
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so we obtain a (functorial and strict!) isomorphism between cones: $$$$\mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \cong C(F(f))$$$$ as cochain complexes. Rearranging shifts, we obtain: $$$$\mathrm{Nat}_{\text{dg}}(C(h_f),F) \cong C(F(f))[-1].$$$$ Now, I'd like to compare $$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?

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.
• Thanks! What about if I require $\mathcal A$ to be a dg-category over a field? Aug 7 '14 at 17:13
• Take $R$ to be a $k$-algebra... Aug 7 '14 at 20:16