In Lemma 3.8 of The Riemann Roch theorem and Serre duality, p.4, it is stated that the refinement map $H^1(\mathfrak U, \mathscr F)\rightarrow H^1(\mathfrak V,\mathscr F)$ is injective, where $\mathfrak U=(U_i)_{i\in I}$ and $\mathfrak V=(V_k)_{k\in K}$ and $\mathfrak V$ is finer than $\mathfrak U$. But I don't see why this is true.
This means if there exists $g_k\in \mathscr F(V_k)$ such that $f_{\tau(k)\tau(l)}\mid_{V_{kl}}=g_l\mid_{V_{kl}}-g_k\mid_{V_{kl}}$, then $(f_{ij})\in Z^1(\mathfrak U,\mathscr F)$ actually lies in $B^1(\mathfrak U,\mathscr F)$. I tried to lift these $g_k$ to elements in $\mathscr F(U_i)$ in order to show that $f_{ij}$ is a coboundary, but it should not be possible, for if these $g_i$ can be pasted together, then we would require $f_{\tau(k)\tau(l)}\mid_{V_{kl}}=0$ for every $k,l$.
I don't find this statement elsewhere, and I run out of ideas, so any help is sincerely appreciated. Thanks in advance.