Exact sequences of sheaves and isomorphisms between them

Question

Suppose we have an exact sequence $0\rightarrow \mathcal{F}'\rightarrow \mathcal{F}\rightarrow \mathcal{F}''\rightarrow 0$ of sheaves, and an exact sequence $0\rightarrow \mathcal{G}'\rightarrow \mathcal{G}\rightarrow \mathcal{G}''\rightarrow 0$ of sheaves with $\mathcal{F}\cong\mathcal{G}$ and $\mathcal{F}''\cong\mathcal{G}''$. Are we then able to conclude that $\mathcal{F}'\cong \mathcal{G}'$?

Answer 1

This question fails at the level of abelian groups. For example, let $F = G = \mathbb Z \times \mathbb Z \times \cdots$. Consider $F' = \mathbb Z$ and $G' = \mathbb Z \times \mathbb Z$. Using the obvious inclusion maps, we note that $F/F' \cong G/G'$. By taking the constant sheaves of these groups, we get the desired exact sequences of sheaves. Clearly, $F' \not \cong G'$ and hence $\mathcal F' \not \cong \mathcal G'$.

Answer 2

No. For instance, consider $\mathbb{P}^1$ with coordinates $(x,y)$ and two morphisms $$ f, g \colon \mathcal{O} \oplus \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}(2), $$ where $$ f = (x^2,xy,y^2), \qquad \text{and} \qquad g = (x^2,0,y^2). $$ Both are surjective, but $$ \operatorname{Ker}(f) \cong \mathcal{O}(-1) \oplus \mathcal{O}(-1), \qquad \text{while} \qquad \operatorname{Ker}(g) \cong \mathcal{O} \oplus \mathcal{O}(-2). $$