Endomorphisms in an exact sequence of vector bundles

Question

Let X be a smooth projective variety over $\mathbb{C}$. And suppose we have an exact sequence of vector bundles over $X$.

$\qquad\qquad\qquad\qquad\qquad 0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$.

Suppose $C$ is simple, i.e. $End(C)=\mathbb{C}$, is $A$ simple too?

My argument is as follows. Since $A$ is a subsheaf of $B$, $End(A)\subset End(B)$. Choose $\phi\in End(A) \subset End{B}$. So $\phi$ is a morphism from $B\longrightarrow B$, such that $\phi(A)\subset A$. Hence it induces a morphism $C\longrightarrow C$. Hence $\phi\in\mathbb{C}$.

Is this argument correct? The step where I have a doubt is that $End{A}\subset End{B}$. Any feedback will be appreciated!

Thanks in advance!

Answer 1

Here's a counterexample. Let $X = \mathbb{P}^1$. Take $C = \mathcal{O}(2)$, which has three global sections, i.e. there is a surjective map $\mathcal{O}^{\oplus 3} \to \mathcal{O}(2)$, which is an isomorphism on global sections. Let $K$ be the kernel, a rank-2 vector bundle.

Since every vector bundle on on $\mathbb{P}^1$ splits as a sum of line bundles, $K = \mathcal{O}(a) \oplus \mathcal{O}(b)$. The Chern class of $K$ must satisfy

$$c(K) = (1+a[pt])(1+b[pt]) = \frac{1}{1+2[pt]} = 1 - 2[pt],$$

so $a + b = -2$. On the other hand, $K$ has no global sections, by construction, so $a,b\leq -1$. So $a = b = -1$ (incidentally, it's probably easy to see this directly, I'm just being lazy), and so $End(K) = \mathbb{C}^4$.

Moreover, in this example, $Hom(B,A) = Hom(\mathcal{O}^{\oplus 3},\mathcal{O}(-1)^{\oplus2}) = 0$.

Answer 2

If $0\to A\to B\to C\to 0$ is exact, so is $0\to A\oplus A\to B\oplus A\to C\to 0$, where the second copy of $A$ maps as identity to the same factor in the middle. Surely, $A\oplus A$ is not simple.