I was doing professor Vakil's FOAG, in exercise 7.2D needs to prove:
Given a morphism of ringed spaces $\pi: X \rightarrow Y$ with $\pi(p)=q$, show that there is a map of stalks $\left(\mathscr{O}_{\mathrm{Y}}\right)_{\mathrm{q}} \rightarrow\left(\mathscr{O}_{\mathrm{X}}\right)_{\mathrm{p}}$.
My attempt, we have the inclusion $i_1:\{p\}\to X, i_2: \{q\}\to Y$. Then we have the inverse image sheaf $i_1^{-1}\mathcal{O}_X, i_2^{-1}\mathcal{O}_Y$ therefore we can define the ringed space $(\{p\},i_1^{-1}\mathcal{O}_X),(\{q\},i_2^{-1}\mathcal{O}_Y)$ and the commutative diagram:
$\require{AMScd}$ \begin{CD} \{p\} @>{\pi}>> \{q\}\\ @VVV @VVV\\ X @>{\pi}>> Y \end{CD}
And since $X\to Y$ is a morphism of ringed space, it has $\pi^\sharp:\pi^{-1}\mathcal{O}_Y\to \mathcal{O}_X$ , and since $i_1^{-1}$ is a functor apply on both side we have $i_1^{-1}\pi^\sharp:i_1^{-1}\pi^{-1}\mathcal{O}_Y\to i^{-1}_1\mathcal{O}_X$ this is the map we want by taking stalk on $\{p\}$ ( where we need to use that $i_2\circ\pi = \pi\circ i_1$) .
Is my proof correct?
The stalk at $p$ is defined to be $\operatorname{colim}_{p\in U}\mathcal{O}_X(U)$ taken over the open sets containing $p$, where if $p\in U\subseteq U'$ we have a restriction homomorphism $\mathcal{O}_X(U')\to\mathcal{O}_X(U)$ and take the colimit along these. Similarly the stalk at $q$ is $\operatorname{colim}_{q\in V}\mathcal{O}_Y(V)$.
Now for each open $V$ containing $q$ there is a homomorphism $\mathcal{O}_Y(V)\to\mathcal{O}_X(\pi^{-1}V)\to(\mathcal{O}_X)_p$ and by the universal property of colimits there is a unique homomorphism $(\mathcal{O}_Y)_q=\operatorname{colim}_{q\in V}\mathcal{O}_Y(V)\to(\mathcal{O}_X)_p$.