I am reading a proof of the following statement, which I cannot understand how it follows.
Proposition: The kernel of every morphism of $\mathcal O_X-$modules $L:\mathcal O_X^a\to \mathcal O_X^b$ is locally free of rank $\geq a-b$.
$\mathcal O_X$ denotes the sheaf of holomorphic functions over a Riemann Surface $X$.
The proof goes by induction in $b$. The case $b=1$ has been treated before and I am ok with that.
Let $L':\mathcal O_X^a\to\mathcal O_X^{b-1}$ be the map $L'=pL$ with $p$ the projection over the first $b-1$ components. By induction hypothesis, we have $\mathcal K:=\text{Ker}(L)$ is locally free of rank $\geq a+1-b$.
It turns out that Ker$(L)$ is the kernel of the morphism $L'':\mathcal K\to\mathcal O_X$ defined by $L''=qL$ with $q$ being the projection over the last coordinate. I understand this, it is like going to $0$ in two steps.
Now the thing is, how does the proposition follow from this? I guess I have to use that $\mathcal K$ is locally free in any way, but I do not see how.