Properties of stalks, only using the categorical definition

Question

I am taking some time to reflect on the slogan "the universal property encapsulates all you need to know"; any reference or comment on this is very welcome.

Let $x$ be a point of a topological space $X$. Let $\mathcal F$ be a presheaf (or a sheaf if need be) on $X$.

Assuming that

• $\mathcal F$ takes values in a category where elements make sense
• the stalk $\mathcal F_x$ is by definition the colimit on the filtered/directed category indexed by open neighbourhoods $U$ of $x$,

can we easily infer from the universal property that

1. any element of $\mathcal F_x$ comes from one of the $\mathcal F(U)$
2. two "functions" have the same stalk at $x$ if and only if their restrictions to a common open set $U$ coincide

without knowing the usual construction?

I was only able to find on page 5 of Sheaf theory, Tennison that the converse is true (a cocone with those properties will be a universal cocone) in the category of sets..

Thank you.

Answer 1

This is more to be seen as a hint: For a sheaf of sets the answer to your first question is particularly easy: assume there is a element in $f \in \mathcal{F}_x$ that doesnt come from some $\mathcal{F}(U)$ and now take $ \mathcal{F}_x \cup \{1\}$ where $1$ should be seen as an arbitrary element that wasn't already in $\mathcal{F}_x$ now proof that $ \mathcal{F}_x \cup \{1\}$ still commutes with all maps but there are at least two maps from $ \mathcal{F}_x \to \mathcal{F}_x \cup \{1\}$ .

For the 2. question the if direction is easy, for only if assume there is no such open set then all the maps $\mathcal{F}(U)\to \mathcal{F}_x$ are not injective (for $U$ small enough) and you can construct "another" stalk $\mathcal{F}'_x$ which has one element more. Suppose $f,g$ are the elements in question. We know that they get maped to the same element in $\mathcal{F}_x$ under the maps $\varphi_U$ (here I assume that $U\subset V\cap W$ where $V,W$ are the sets where $f,g$ come from). Now define the maps $\varphi'_U :\mathcal{F}(U)\to \mathcal{F}_x\cup \{*\} $ by

$$ \varphi'_U(x)= \begin{cases} * \qquad \text{ if } x \in res_{U',U}^{-1}(g_{\vert U'}) \text{ for some } U'\subset U\\ \varphi_U(x)\qquad \text{else } \\\end{cases}. $$

Now a map $\alpha: \mathcal{F}_x\to \mathcal{F}_x\cup \{*\}$ with the property $\alpha\circ \varphi_U=\varphi'_U$ cant exist because $f,g$ get send to the same element by $\varphi_U$ and therefore by $\alpha\circ \varphi_U$ but to different elements by $\varphi_U'$.