Let $(X,\mathcal{O}_X)$ be an affine scheme and $\mathcal{F}$ a quasi-coherent sheaf. Then for any inclusion $j:U\hookrightarrow X$ for an open $U$, is the natural map $$\operatorname{Hom}(\mathcal{F},\mathcal{F})\rightarrow \operatorname{Hom}(\mathcal{F}|_U,\mathcal{F}|_U)$$ surjective?
Intuitively, it feels so, but I'm worried that this may fail possibly as you have to be careful when extending a morphism of sheaves...
If this is true, is it also true in more generality, that is for a ringed space and any sheaf of Abelian groups?
CONEXT
This popped up while I was trying to show that
The natural morphism (given by the adjointness) $$\mathcal{F}\rightarrow j_{\ast}(\mathcal{F}|_U)$$ is a monomorphism of sheaves.
EDIT I had made a mistake in my context: it should rather be
$\mathcal{F}\rightarrow j_\ast j^\ast\mathcal{F}$ is injective iff $\operatorname{Hom}(\mathcal{F},\mathcal{F})\rightarrow \operatorname{Hom}(\mathcal{F},j_\ast j^\ast\mathcal{F})$ is injective
This is true iff $\operatorname{Hom}(\mathcal{F},\mathcal{F})\rightarrow \operatorname{Hom}(j^\ast\mathcal{F}, j^\ast\mathcal{F})$ is injective by adjointness.
So the claim I would rather what to prove is, with the assumptions above,
Is the natural map $$\operatorname{Hom}(\mathcal{F},\mathcal{F})\rightarrow \operatorname{Hom}(\mathcal{F}|_U,\mathcal{F}|_U)$$ injective?
Again this feels like its bound to fail with a good enough counter-example.
Any help will be appreciated! Thank you!
Original answer, before question was changed: No, this is very badly not true. Here's one simple example: consider $\mathcal{O}_X$. As $\operatorname{Hom}(\mathcal{O}_X,\mathcal{F})\cong\mathcal{F}(X)$ for any $\mathcal{F}$, we find that $\operatorname{Hom}(\mathcal{O}_X,\mathcal{O}_X)\cong\mathcal{O}_X(X)$, while $\operatorname{Hom}(\mathcal{O}_X|_U,\mathcal{O}_X|_U)\cong\mathcal{O}_X(U)$. So if your claim was correct, $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ would always be a surjection for any open $U$. On the other hand, consider $D(t)\subset X=\operatorname{Spec} k[t]$, where $\mathcal{O}_X(X)=k[t]$ and $\mathcal{O}_X(D(t))=k[t,t^{-1}]$, and there's no surjection here.
After the question was changed: let me get straight to the point with your underlying question. Here is a counterexample:
Let $X=\operatorname{Spec} k[t]$, let $\mathcal{F}=\widetilde{k[t]/(t)}$, and let $U=D(t)$. Then $j^*\mathcal{F}$ is the sheaf associated to the $k[t,t^{-1}]$-module $k[t]/(t)\otimes_{k[t]} k[t,t^{-1}]$, which is zero, so $j^*\mathcal{F}\cong 0$. Thus $\mathcal{F}\to j_*j^*\mathcal{F}$ is the zero map but $\mathcal{F}$ is not the zero sheaf, and so this map is not a monomorphism.