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Let $\mathcal{A}$ be an abelian category and $\mathcal{B}\subseteq \operatorname{Proj}(\mathcal{A})$ be a full additive subcategory of $\mathcal{A}$. We define the full subcategory $\mathcal{B(A)}$ of $\mathcal{A}$ whose objects are those $A\in \mathcal{A}$ such that there exists an exact sequence $$B_1\to B_0\to A\to 0,\quad \text{with}\quad B_i\in\mathcal{B}.$$ Note that $\mathcal{B}\subseteq\mathcal{B(A)}$. In Proposition (2.1) of Coherent Functors, M. Auslander constructs a left adjoint for the functor $J\colon \operatorname{Fun}(\mathcal{B(A)},\mathcal{D})\to \operatorname{Fun}(\mathcal{B},\mathcal{D})$ induced by the inclusion functor $\mathcal{B}\hookrightarrow \mathcal{B(A)}$, where $\mathcal{D}$ is an additive category with cokernels.

His proof is as following: the candidate for being the left adjoint is the functor $$L\colon\operatorname{Fun}(\mathcal{B},\mathcal{D})\to \operatorname{Fun}(\mathcal{B(A)},\mathcal{D})$$ defined on functors $F\colon \mathcal{B}\to\mathcal{D}$ by the following data: for every object $X\in\mathcal{B(A)}$ let be fixed an exact sequence $\eta_{X}\colon X_1\to X_0\to X\to 0$; and if $X\in\mathcal{B}$, set $\eta_{X}\colon 0\to X\xrightarrow{1_X}X\to 0$. Then, $$LF(X)\colon = \operatorname{Coker}(FX_1\to FX_0).$$ Note that $LF(X)=FX$, for $X\in\mathcal{B}$. On morphisms $f\colon X\to Y$ in $\mathcal{B(A)}$, using the fact that objects in $\mathcal{B}$ are projectives, $LF(f)\colon LF(X)\to LF(Y)$ is the morphism induced by the universal property of cokernels. Now, there is only a way to define the image $L(\alpha)\colon LF\to LF'$ of a natural transformation $\alpha\colon F\to F'$. Finally, Auslander says that by restricting the components, the map $$\tau_{F,G}\colon \alpha \mapsto \alpha\mid_{\mathcal{B}},$$ is an isomorphism from $\operatorname{Nat}(LF,G)\to \operatorname{Nat}(F,JG)$. Clearly, it's well-defined. But my question is: why is this map an isomorphism?

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  • $\begingroup$ I think given an arbitrary natural transformation (nt) $\tau:F\Rightarrow JG$, we construct the corresponding nt $LF\Rightarrow G$ as follows: for any $X\in\mathcal{B}(\mathcal{A})$, apply $LF$ and $G$ to $\eta_X$ and the resulting diagrams are now connected by $\tau$ for $X_1$ and $X_0$ in $\eta_X$ and then use the universality of cokernel $LF(X)$ to construct the component $LF(X)\rightarrow G(X)$. $\endgroup$
    – Sampah
    Commented Aug 29, 2023 at 8:55
  • $\begingroup$ @Sampah but why universality of cokernel applies? Because $G(X_1\to X_0\to X)=G(0)$ is not necessarily zero. $\endgroup$ Commented Aug 29, 2023 at 14:27
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    $\begingroup$ @Sampah no, but you made me check and now I found that Auslander is assuming that all functors are additive and that suffices... my bad. $\endgroup$ Commented Aug 29, 2023 at 14:46
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    $\begingroup$ Ah so the functors are taken to be right exact (makes sense) and additive. No worries glad you have solved it. $\endgroup$
    – Sampah
    Commented Aug 29, 2023 at 14:50
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    $\begingroup$ @Sampah just merely additive, right exactness of $G$ is not required. When $G$ is right exact, $G\cong (LJ)G$, and the converse is also true. $\endgroup$ Commented Aug 29, 2023 at 15:02

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