Let $i : U \to X$ be an open inclusion. Is $ i_* i^{-1} $ an exact functor?
I think possibly not, but I am kind of convinced yes (because the maps on stalks are either the same or made into zero).
The reason why I think it is not is because I am reading about sheaf cohomology, and there is the following:
$F$ is a sheaf, embeded in an injective sheaf $I$, giving an exact sequence:
$0 \to F \to I \to C \to 0$.
The author then restricts via $i_* i^{-1}$, notation: $i_* i^{-1} F = F_j$, to get a sequence
$0 \to F_j \to I_j \to \bar{C_j} \to 0$, where $\bar{C_j}$ is not the restriction of $C$, but rather the quotient sheaf of $I_j / F_j$.
The author seems to suggest that this quotient $\bar{C_j}$ is a proper subsheaf of $C_j$ (the restriction of $C$). I do not understand why it should be a proper subsheaf, since it feels to me that $i_* i^{-1}$ should be exact in this situation.
Is there a good example to think about?
Later on he proves in a stronger circumstance that $\bar{C_j} = C_j$.
(This is on the course to proving Serre's vanishing theorem about the cohomology of quasi-coherent sheaves on affine schemes.)
In general, the functor $i_*:Sh(U)\rightarrow Sh(X)$ is not exact. But before I'll give some examples, let me point out that this is not true that the stalk of $i_*i^{-1}\mathcal{F}$ at $x$ is either $\mathcal{F}_x$ or $0$ depending on whether $x\in U$ or not. There is the so called extension by zero functor $i_!$ which has this property.
Indeed, take $X=\mathbb{R}$ and $U=\mathbb{R}\setminus\{0\}$. Then $(i_*\mathbb{Z})_0=\mathbb{Z}\oplus\mathbb{Z}$. This is because for any small enough neighborhood $V$ of $0$, $V\cap U$ has two connected components.
The functor $i_*$ is not exact, and it has higher direct images. For example, on $X=\mathbb{C}$ and $U=\mathbb{C}\setminus\{0\}$, $R^1i_*i^{-1}\mathbb{Z}=\mathbb{Z}_0$ is a skyscraper sheaf placed at the origin. This is because for any small enough neighborhood $V$ of $0$, $V\cap U$ is homotopically a circle.
This gives an example where the functor $i_*i^{-1}$ is not exact : consider the exponential sequence on $X=\mathbb{C}$ : $$0\rightarrow\mathbb{Z}\rightarrow\mathcal{O}\overset{\exp}\rightarrow\mathcal{O}^\times\rightarrow 0$$ where $\mathcal{O}$ is the sheaf of holomorphic functions on $\mathbb{C}$ and $\mathcal{O}^\times$ the sheaf of non vanishing holomorphic functions. If you apply $i_*i^{-1}$ you get : $$0\rightarrow\mathbb{Z}\rightarrow i_*\mathcal{O}\rightarrow i_*\mathcal{O}^\times\rightarrow\mathbb{Z}_0\rightarrow 0$$ This sequence means that a non vanishing holomorphic function $f$ defined on $V\setminus\{0\}$ where $V$ is a neighborhood of $0$ has a logarithm if and only if the residue of $f'/f$ at $0$ is zero.