Induced morphism on stalks is well defined

Question

Let $\pi:X\rightarrow Y$ be a morphism of ringed spaced with $\pi(p)=q$. We then have a map $\pi^\#:\mathcal{O}_Y\rightarrow \pi_*\mathcal{O}_X$ of the structure sheaves. It makes sense that our map of stalks $(\mathcal{O}_Y)_p\rightarrow (\mathcal{O}_x)_p$ should be given by $[(V,s)]\mapsto [(\pi^{-1}(V),\pi^\#_V(s))]$.

But I'm having trouble seeing why this map is well defined. Suppose $[(V,s)]=[(U,t)]$. Then there exists some open $W\subset U,V$ such that $s|_W=t|_W$. But then does this imply that $[(\pi^{-1}(V),\pi^\#_V(s))] = [(\pi^{-1}(U),\pi^\#_U(s))]$? I know in principal I want to restrict the sections $\pi^\#_V(s)$ and $\pi^\#_V(s)$ to some open subset (probably $\pi^{-1}(W)$) and check that they are equal, but I am having trouble doing this. What am I missing?

Answer 1

By definition of morphism of sheaves and pushforward of sheaves, this diagram commutes:

$\newcommand{\O}{\mathcal{O}}$ $\require{AMScd}$ \begin{CD} \O_Y(U) @>f^\#_U>> \O_X(f^{-1}(U)) \\ @V res V V @VV res V\\ \O_Y(W) @>>f^\#_W> \O_X(f^{-1}(W)) \end{CD}

Therefore, for $s \in \O_Y(U)$, $f^\#_U(s)|_{f^{-1}(W)} = f^\#_W(s|_W)$, and similarly for $t \in \O_Y(V)$, $f^\#_V(t)|_{f^{-1}(W)} = f^\#_W(t|_W)$. Since $s|_W = t|_W$, their image under $f^\#_W$ are also the same. Therefore, $f^\#_U(s)$ and $f^\#_V(t)$ restrict to the same thing in $f^{-1}(W)$

Answer 2

It’s abstract nonsense (the most complex kind of nonsense), really. Indeed, $\pi_V^{\sharp}(s)_{|\pi^{-1}(W)}=\pi_W^{\sharp}(s_{|W})=\pi_W^{\sharp}(t_{|W})=\pi_U^{\sharp}(t)_{|\pi^{-1}(W)}$.