Let us denote by $i:U\rightarrow X$ an open inclusion and $\mathcal{F}$ a coherent sheaf on $U$.


Is the map $i\_!\mathcal{F} \rightarrow i\_\*\mathcal{F}$ always injective ?


I ask this question because it seemed to me that we define the minimal extension $i\_{!\*}\mathcal{F}$ as the image of $i\_!\mathcal{F} \rightarrow i\_\*\mathcal{F}$, and if we want to get an interesting extension (not the extension by zero), it would be better not to have $i\_{!\*}\mathcal{F} \simeq i\_{!}\mathcal{F}$.


Probably I'm confuse and I don't see an obvious thing, but an answer of Donu Arapura [here](https://mathoverflow.net/questions/304773/projective-direct-image-and-commutation), concerning a question of commutation I had, makes me doubt.


Thank you. Jean.



Proper direct image and minimal extension
Here’s what your misunderstanding: $i\_{!\*}\mathcal F$ is the image of $i\_!\mathcal F \to i\_\* \mathcal F$ *in the category of perverse sheaves*. It’s instructive to look at the case $U=\Bbb C^\*$ and $X=\Bbb C$. Try showing that the cone of $$i\_! \Bbb C\_U[1] \to i\_\* \Bbb C\_U[1]$$ is not perverse (in the middle perversity). Note that I’m assuming that you’re using $i\_\*$ for the derived direct image functor. 


