Kernel,image of quasi-coherent sheaf is quasi-coherent for ringed spaces?

Question

If $X$ is a ringed or locally ring space (not necessarily scheme), do we still have the kernel, image of quasi-coherent sheaves quasi-coherent?

Answer 1

How about this for kernels? As a topological space let $X$ be the real line. As our sheaf of regular functions on an open $U$ we take continuous functions $f : U \to \mathbf{C}$ which are given by a polynomial on $U \cap (-\infty, \epsilon)$ for some small $\epsilon > 0$. In other words, the structure sheaf strictly to the right of $0$ is given by continuous functions and the structure sheaf on $(-\infty, 0]$ is given by polynomial functions.

For $n \geq 1$ let $f_n(x) = \max(0, x - 1/n)$. Consider the map $$ (f_1, f_2, f_3, \ldots) : \mathcal{O}_X \longrightarrow \bigoplus\nolimits_{n \geq 1} \mathcal{O}_X $$ Edit: Does not work, because $(f_1, f_2, \ldots)$ is not a section of the direct sum. See comment of OP below.

I claim the kernel $\mathcal{K}$ is not quasi-coherent. Namely, any section over an open $U$ of the kernel is zero on $U \cap (0, \infty)$. Hence by our construction of $\mathcal{O}_X$ we see that the stalk of the kernel at $0$ is zero. This would imply that $\mathcal{K}$ if it were even generated by local sections around $0$, would be zero in a neighbourhood of $0$.