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Mar 11, 2021 at 1:12 comment added alg_et_geom I see, thanks for your help!
Mar 10, 2021 at 23:29 comment added R. van Dobben de Bruyn Ah, that's certainly true, and follows from the stated version together with the version in my first comment. Indeed, if $0 \to \mathscr G_+ \to \mathscr G \to \mathscr G' \to 0$ is the semistable subobject of highest slope (for whatever slope function you're using), then $\operatorname{Hom}(\mathscr F, \mathscr G') = 0$ so $\psi$ lands in $\mathscr G_+$, which is semistable with $p(\mathscr G_+) = p(\mathscr F)$.
Mar 10, 2021 at 21:24 comment added alg_et_geom @R.vanDobbendeBruyn Thanks for your comments. I suppose my question is whether there's a statement like: $p_-(F)=p_+(G)$ and $F$ stable implies $\psi$ is injective or zero? Because if this were true, then instead of requiring $p(F)=p(G)$ for injectivity, I suppose we could require $p(F)=p_+(G)$
Mar 10, 2021 at 20:41 comment added R. van Dobben de Bruyn By contrast, if $\mathscr F$ and $\mathscr G$ are stable but $p(\mathscr F) < p(\mathscr G)$, then there are many nonzero maps $\mathscr F \to \mathscr G$, which may even have kernels. For example, if $\mathscr F$ is stable of rank $2$ and $s \in H^0(\mathscr F^\vee(d))$ for $d \gg 0$, then $s$ gives a map $\mathscr F \to \mathcal O(d)$ whose kernel must be a line bundle.
Mar 10, 2021 at 20:35 comment added R. van Dobben de Bruyn When $\mathscr F$ is stable (or even semistable), we have $p_-(\mathscr F) = p(\mathscr F) = p_+(\mathscr F)$, so $p(\mathscr F) > p_+(\mathscr G)$ implies $\psi = 0$. I don't know what you mean by "there doesn't seem to be a statement regarding injectivity" ― the conclusion is actually stronger in the case of different slopes. (Compare this with Schur's lemma for representations of finite groups, where there are no nonzero maps between different irreducible characters.)
Mar 10, 2021 at 19:36 history asked alg_et_geom CC BY-SA 4.0