Let $ X $ be a topological space. Let $ P $ and $ Q $ be two presheaves on $ X $, and let $ h\colon P\to Q $ be a morphism of presheaves. Let $ P_x $ be the stalk of $ P $ at $ x\in X $ and the same $ Q_x $. Let $ h_x\colon P_x\to Q_x $ be the unique map that makes $$ \require{AMScd} \begin{CD} PU @>>> P_x\\ @V{h_U}VV @VV{h_x}V\\ QU @>>> Q_x \end{CD} $$ commute for every $ U\subset X $ open, where the horizontal arrows maps a section to its germ. Let $$ \Lambda P\colon \coprod_{x\in X}P_x\to X $$ be the obvious projection map $ (x,\mathbf g)\mapsto x $, and the same for $ \Lambda Q $. Finally, for $ U\subset X $ open and $ s\in PU $ let $$ \underline s\colon U\to \coprod_{x\in X}P_x $$ denote the map $ s\mapsto (x,s_x) $, and the same for any $ t\in QU $.
I'm trying to show that the coproduct map $$ \Lambda h = \coprod_{x\in X}h_x\colon \coprod_{x\in X}P_x\to \coprod_{x\in X}Q_x $$ is continuous, where disjoint union is equipped with the final topology with respect to the maps $ U\to \coprod_{x\in X}P_x $ or $ U\to \coprod_{x\in X}Q_x $ introduce above, but I do not have succeeded yet. Could someone prove me any hint?
Ok, I think I've come to a solution. It wasn't that difficult in the end. The composite map $$ U\xrightarrow{\underline s} \coprod_{x\in X}P_x\xrightarrow{\Lambda h} X $$ is nothing but the map $ \underline{h_U(s)}\colon U\to \coprod_{x\in X}Q_x $.