I am reading (out of personal interest/self-study) the chapter on sheaves in Warner Foundations of Differentiable Manifolds and Lie Groups book and there is a point I do not understand in his proof of the fact that the module operations in the sheaf constructed out of a presheaf are continuous. This is on pages 166-167 of the book. I have tried to make the question self-contained.
Following his notation, a set $\mathcal{S}\xrightarrow{\pi} M$ has been constructed out of a presheaf via the direct limit construction. The topology on $\mathcal{S}$ is the one induced by the basis \begin{equation} O_f = \{\rho_{p,U}\ f : p\in U \} \end{equation} as $f\in S_U$, the module associated to $U$ by the presheaf, and $U$ varies among the open sets in $M$. The map $\rho_{p,U}$ projects $f$ to its equivalence class in the direct limit construction (the equivalence relation being agreeing on an open subset up to taking restriction). At this point of the proof we know that $\pi$ is a surjective continuous open map (in fact a local homeomorphism), hence in particular a quotient map.
While proving that for $m\in M$, $s_i\in\pi^{-1}(m)$, the map $(s_1,s_2)\mapsto s_1-s_2$ is continuous (with the product topology on $\mathcal{S}\times\mathcal{S}$), Warner constructs a set of the form \begin{equation} O_{\rho_{Q,V}\ \ \ g}\times O_{\rho_{Q,W}\ \ \ h} \cap \mathcal{S}\circ \mathcal{S}, \end{equation} where $\mathcal{S}\circ\mathcal{S}$ is the susbspace of $\mathcal{S} \times \mathcal{S}$ given by elements having the same image under $\pi$, which he claims to be open.
Now $O_{\rho_{Q,V}\ \ \ g}\times O_{\rho_{Q,W}\ \ \ h}$ is open (in the product topology) as the product of basic open sets. However, it seems to me that $\mathcal{S}\circ\mathcal{S}$ is closed. In fact $M$ is assumed to be Hausdorff, and $\pi$ is an open quotient map which I think implies that $\mathcal{S}\circ\mathcal{S} $ is closed. In which case I don't see why $O_{\rho_{Q,V}\ \ \ g}\times O_{\rho_{Q,W}\ \ \ h} \cap \mathcal{S}\circ \mathcal{S}$ should be open.
I have figured out the answer while writing, but since this might be useful to someone else I thought I could still post the question and also write an answer. Feel free to delete should this be frowned upon.
While I assumed that, for $\mathcal{S}$ a sheaf, the map $\mathcal{S}\times \mathcal{S} \rightarrow \mathcal{S}$ given by $(s_1,s_2)\mapsto s_1-s_2$ has to be continuous with respect to the product topology on $\mathcal{S}\times \mathcal{S} $, that is not correct.
The operation itself is only defined for points on the same stalk, and continuity is, by definition see page 163, requested for the map on $\mathcal{S}\circ \mathcal{S} \rightarrow \mathcal{S}$, where I imagine that $\mathcal{S}\circ \mathcal{S} $ has the topology as a subspace of $\mathcal{S}\times \mathcal{S} $. Hence there are no problems with the proof.