What is a natural morphism of stalks?

Question

I am a bit confused with this question. Let $\pi: X \rightarrow Y$ be a continuous map and $F$ a sheaf of sets on $X$. If $\pi(p) =q$, then what is the natural morphism of stalks $(\pi_*F)_q \rightarrow F_p$, where $\pi_*$ is the pushforward of $F$ by $\pi$? I would appreciate any assistance. Thanks!

Answer 1

By definition of a direct image,

$$\pi_*(F)(U)=F(\pi^{-1}(U))$$

the stalk is the limit of a system

$$\pi_*(F)_q=\lim\limits_{q \in U}\pi_*(F)(U) =\lim\limits_{q \in U}F(\pi^{-1}(U))$$ Now if $q \in U$ then $p\in \pi^{-1}(U)$ so the system which limits to $\pi_*(F)_q$ is a subsystem of that which limits to $F_p$, this gives a natural map.

Answer 2

This is a general property of limits and colimits (and probably even more general things). I'll demonstrate in the case of equalizers to give you the general idea.

Suppose you have a pair of arrows $f,g : A \to B$ and another pair of arrows $f',g' : A' \to B'$ and you have a morphism from the first diagram to the second: that is, you have arrows

$$ u : A \to A' \qquad \qquad v : B \to B' $$

such that you have the appropriate commutative squares: i.e. $vf = f'u$ and $vg = g'u$.

Suppose $e : E \to A$ is an equalizer of $f$ and $g$ and $e' : E' \to A'$ is an equalizer of $f'$ and $g'$.

The arrow $ue$ has the property that $f'ue = g'ue$, therefore there exists a unique arrow $w:E \to E'$ with the property that $ue = e'w$.

The arrow $w$ is the "natural morphism" between the two equalizers that is induced by the morphism of diagrams.

Answer 3

Given that the stalk is a filtered colimit, take $F_p$ as a cone over the diagram $I \to (\pi_\ast F)$ – which can be done because for all open $V \subset Y$, $\pi^{-1}(V)$ is an open subset of $X$ and $F_p$ being a colimit of open subsets of $X$ containing $p$ has maps into it from each $\pi^{-1}(V)$. Thus by the universal property characterizing the colimit there will exist a unique map $(\pi_\ast F)_q \to F_p$.