A morphism of semi-stable vector bundle of degree $0$ is of constant rank?

Question

Let $V_1$ and $V_2$ be semistable vector bundles of degree $0$ over a smooth projective curve $X$. How does one prove that any $f\in Hom(V_1,V_2)$ has constant rank? That is the rank of the linear maps $f_x:(V_1)_x\rightarrow (V_2)_x$ doesn't change with respect to $x\in X$.

Answer 1

In the title you say degree 0, in the body you don't. Without that assumption, the answer is no.

Assuming degree 0, let $f:V_1\to V_2$ be such a map and let $W=f(V_1)$. Semistability says that $\deg W\leq 0$ and then since the kernel of $f$ also has degree less than or equal to zero, we see that both must be degree 0. If $W$ is not a subbundle, then its saturation in $V_2$ will have positive degree, which is not possible, so $W$ is a subbundle and that precisely says the map has constant rank.