Exactness of $j_!$ in abelian category recollement Consider a recollement situation, with notation the same as on the nLab page. That is, we have adjunctions $i^* \dashv i_* \dashv i^!$ and $j_! \dashv j^* \dashv j_*$ between the abelian categories $\mathsf{A}' \xrightarrow{i_*} \mathsf{A} \xrightarrow{j^*} \mathsf{A}''$ such that

*

*$i_*$, $j_!$ and $j_*$ are fully faithful,


*The essential image of $i_*$ is equivalent to the kernel $\operatorname{Ker}(j^*)$.
Purely from the adjunctions in the recollement and (co)continuity of (left)/right adjoints, we have that


*$i_*$ and $j^*$ are exact,


*$i^*$ and $j_!$ are right exact,


*$i^!$ and $j_*$ are left exact.

Question 1: If $i^*$ is left exact, then is $j_!$ left exact?
Question 2: If $j_!$ is left exact, then is $i^*$ left exact?

Here is the motivation for this question. Consider the specific recollement situation of sheaves of abelian groups on topological spaces for the inclusions $Z \overset{i}{\hookrightarrow} X \overset{j}\hookleftarrow X \setminus Z$ where $Z$ is a closed subspace of $X$.
In this case, we know that the inverse image of sheaves is exact, so in particular $i^*$ is also exact.
In this sheaf recollement situation, $j_!$ is usually proved to be left exact using a specific construction of the functor. For example, the exactness of $j_!$ is stated in proposition 5.4.2 of Etale Cohomology Theory by Lei Fu.
I am wondering if the exactness of $j_!$ here follows only from "abstract nonsense," the recollement axioms, and exactness of $i^*$, or if we actually need to construct the functor $j_!$ in order to verify its exactness?
 A: The answer to Question 1 is "yes".
I'll identify $\mathsf{A}'$ with a Serre subcategory of $\mathsf{A}$ via $i_*$, and the quotient $\mathsf{A}/\mathsf{A}'$ with $\mathsf{A}''$ via $j^*$.
Then for any $X$ in $\mathsf{A}$, $i^*X$ is the largest quotient of $X$ that is in $\mathsf{A}'$. If $i^*$ is left exact, then the kernel of the quotient map $X\to i^*X$ has no nonzero subquotient in $\mathsf{A}'$, since if $Y$ were a subobject of the kernel with a nonzero quotient in $\mathsf{A}'$, then the inclusion $Y\to X$ would be a monomorphism that is sent to the zero map $0\neq i^*Y\to i^*X$ by $i^*$.
Let $\mathsf{B}$ be the Serre subcategory of $\mathsf{A}$ consisting of objects with no nonzero subquotient in $\mathsf{A}'$. Then the functor $\mathsf{A}\to\mathsf{B}$ sending $X$ to the kernel of $X\to i^*X$ is a quotient functor with kernel $\mathsf{A}'$. So we can identify $\mathsf{B}$ with $\mathsf{A}/\mathsf{A}'$, and then the left adjoint $j_!$ of the quotient functor is just the inclusion $\mathsf{B}\to\mathsf{A}$, which is exact.
However, the answer to Question 2 is "no".
For example, there is a recollement where $\mathsf{A}'$ and $\mathsf{A}''$ are both the category of abelian groups, and $\mathsf{A}$ is the category of morphisms of abelian groups (i.e., an object of $\mathsf{A}$ is a diagram $[G_1\to G_2]$ of abelian groups), with $i_*G=[0\to G]$ and $j^*[G_1\to G_2]=G_1$.
The left adjoints are given by $i^*[G_1\stackrel{\alpha}{\to}G_2]=\text{cok }\alpha$, which is not exact, and $j_!G=[G\stackrel{\text{id}}{\to}G]$, which is exact.
