Complete proof for sanity check for pseudo-semistabs

main.tex

@@ -235,8 +235,22 @@ $\bddderived(X)$, some other sources may have this extra restriction too.
 	Since all the objects in the sequence are in $\firsttilt\beta$, we have
 	$\chern_1^{\beta} \geq 0$ for each of them. Due to additivity
 	($\chern(F) = \chern(E) + \chern(G)$), we can deduce
-	$0 \leq \chern(E) \leq \chern(F)$.
-	% TODO unfinished
+	$0 \leq \chern_1^{\beta}(E) \leq \chern_1^{\beta}(F)$.
+
+
+	$E \hookrightarrow F \twoheadrightarrow G$ being a semistabilizing sequence
+	means	$\nu_{\alpha,\beta}(E) = \nu_{\alpha,\beta}(F) = \nu_{\alpha,\beta}(F)$.
+	% MAYBE: justify this harder
+	But also, that this is an instance of $F$ being semistable, so $E$ must also
+	be semistable
+	(otherwise the destabilizing subobject would also destabilize $F$).
+	Similarly $G$ must also be semistable too.
+	$E$ and $G$ being semistable implies they also satisfy the Bogomolov
+	inequalities:
+	% TODO ref Bogomolov inequalities for tilt stability
+	$\Delta(E), \Delta(G) \geq 0$.
+	Expressing this in terms of Chern characters for $E$ and $F$ gives:
+	$\Delta(\chern(E)) \geq 0$ and $\Delta(\chern(F)-\chern(E)) \geq 0$.
 \end{proof}