Consider a [recollement](https://ncatlab.org/nlab/show/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


1. $i\_\*$, $j\_!$ and $j\_\*$ are fully faithful,
2. 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


3. $i\_\*$ and $j^\*$ are exact,
4. $i^\*$ and $j\_!$ are right exact,
5. $i^!$ 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?



Exactness of $j_!$ in abelian category recollement
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.


