Subsheaves of a direct sum of line bundles

Question

All our sheaves are defined on the projective line. Consider a direct sum sheaf $S=\bigoplus O_{i\in I}(a_i)$ for a finite set $I$. If $S$ is a subsheaf of some $T=\bigoplus_{j\in J} O(1)$ for some finite set $J$, does it follow that $a_i\leqslant 1$? I have been trying to prove this by considering the long exact sequence of cohomology groups, but so far without success. I must be missing something obvious here.

Answer 1

Let's go with your approach. If $\mathcal \oplus_{i \in I} \mathcal O(a_i)$ is a subsheaf of $\oplus_{j \in J}\mathcal O(1)$, then so is $\mathcal O(a_i)$, for each $i\in I$. So for each $i\in I$, we have a short exact sequence, $$ 0 \to \mathcal O(a_i)\to \oplus_{j \in J}\mathcal O(1)\to \mathcal F \to 0.$$ Tensoring everything with $\mathcal O(-a_i)$ (which is locally free, so tensoring preserves exactness), we get $$ 0 \to \mathcal O\to \oplus_{j \in J}\mathcal O(1-a_i)\to \mathcal F(-a_i) \to 0.$$ Now let's look at the LES: $$ 0 \to H^0(\mathcal O) \to \oplus_{j \in J} H^0 (\mathcal O(1 - a_i)) \to H^0( \mathcal F(-a_i)) \to \dots$$ Clearly, $\mathcal H^0(\mathcal O) = k$, where $k$ is the base field.

But if $a_i > 1$, then $H^0(\mathcal O(1-a_i)) = 0$. This is a contradiction.