The Stacks project

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

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.

Thus the lemma holds. $\square$