Lemma 12.7.4. Let $a : \mathcal{A} \to \mathcal{B}$ and $b : \mathcal{B} \to \mathcal{A}$ be functors. Assume that

1. $\mathcal{A}$, $\mathcal{B}$ are additive categories, $a$, $b$ are additive functors, and $a$ is right adjoint to $b$,

2. $\mathcal{B}$ is abelian and $b$ is left exact, and

3. $ba \cong \text{id}_\mathcal {A}$.

Then $\mathcal{A}$ is abelian.

Proof. As $\mathcal{B}$ is abelian we see that all finite limits and colimits exist in $\mathcal{B}$ by Lemma 12.5.5. Since $b$ is a left adjoint we see that $b$ is also right exact and hence exact, see Categories, Lemma 4.24.6. Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$. In particular, if $K = \mathop{\mathrm{Ker}}(B_1 \to B_2)$, then $K$ is the equalizer of $0$ and $\varphi$ and hence $bK$ is the equalizer of $0$ and $b\varphi$, hence $bK$ is the kernel of $b\varphi$. Similarly, if $Q = \mathop{\mathrm{Coker}}(B_1 \to B_2)$, then $Q$ is the coequalizer of $0$ and $\varphi$ and hence $bQ$ is the coequalizer of $0$ and $b\varphi$, hence $bQ$ is the cokernel of $b\varphi$. Thus we see that every morphism of the form $b\varphi$ in $\mathcal{A}$ has a kernel and a cokernel. However, since $ba \cong \text{id}$ we see that every morphism of $\mathcal{A}$ is of this form, and we conclude that kernels and cokernels exist in $\mathcal{A}$. In fact, the argument shows that if $\psi : A_1 \to A_2$ is a morphism then

$\mathop{\mathrm{Ker}}(\psi ) = b\mathop{\mathrm{Ker}}(a\psi ), \quad \text{and}\quad \mathop{\mathrm{Coker}}(\psi ) = b\mathop{\mathrm{Coker}}(a\psi ).$

Now we still have to show that $\mathop{\mathrm{Coim}}(\psi )= \mathop{\mathrm{Im}}(\psi )$. We do this as follows. First note that since $\mathcal{A}$ has kernels and cokernels it has all finite limits and colimits (see proof of Lemma 12.5.5). Hence we see by Categories, Lemma 4.24.6 that $a$ is left exact and hence transforms kernels (=equalizers) into kernels.

\begin{align*} \mathop{\mathrm{Coim}}(\psi ) & = \mathop{\mathrm{Coker}}(\mathop{\mathrm{Ker}}(\psi ) \to A_1) & \text{by definition} \\ & = b\mathop{\mathrm{Coker}}(a(\mathop{\mathrm{Ker}}(\psi ) \to A_1)) & \text{by formula above} \\ & = b\mathop{\mathrm{Coker}}(\mathop{\mathrm{Ker}}(a\psi ) \to aA_1)) & a\text{ preserves kernels} \\ & = b\mathop{\mathrm{Coim}}(a\psi ) & \text{by definition} \\ & = b\mathop{\mathrm{Im}}(a\psi ) & \mathcal{B}\text{ is abelian} \\ & = b\mathop{\mathrm{Ker}}(aA_2 \to \mathop{\mathrm{Coker}}(a\psi )) & \text{by definition} \\ & = \mathop{\mathrm{Ker}}(baA_2 \to b\mathop{\mathrm{Coker}}(a\psi )) & b\text{ preserves kernels} \\ & = \mathop{\mathrm{Ker}}(A_2 \to b\mathop{\mathrm{Coker}}(a\psi )) & ba = \text{id}_\mathcal {A} \\ & = \mathop{\mathrm{Ker}}(A_2 \to \mathop{\mathrm{Coker}}(\psi )) & \text{by formula above} \\ & = \mathop{\mathrm{Im}}(\psi ) & \text{by definition} \end{align*}

Thus the lemma holds. $\square$

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