The following topic shows how the skyscraper sheaf $F_x$ on a pointed space $(X,x)$ with values in the category of sets or groups is defined : A question about the skyscraper sheaf

On the Wikipedia article "Sheaf", they say that the skyscraper sheaf can be further generalised to take values in any category $C$ with a terminal object.

Let's recall the construction :

Fix a point $x$ in $X$ and an object $S$ in a category $C$. The skyscraper sheaf over x with stalk S is the sheaf $S_x$ defined as follows: If U is an open set containing $x$, then $S_x(U) = S$. If $U$ does not contain $x$, then $S_x(U)$ is the terminal object of $C$. The restriction maps are either the identity on S, if both open sets contain $x$, or the unique map from S to the terminal object of $C$.

Indeed, this construction does give a presheaf. But does it really give a sheaf whatever the category $C$ is ? The problem is with the gluing axiom.

Let's call $T$ the terminal object of $C$ and $\phi$ the unique map from $S$ to $T$. Let $\{U_i\}_{i\in I}$ be an open cover of $X$ and, for each $i$, $s_i \in S_x(U_i)$ a section, such that for each pair $U_i,U_j$ such that $U_i \cap U_j \neq \varnothing$ the restrictions of $s_i$ and $s_j$ agree on the overlaps : ${s_i}_{|U_i \cap U_j} = {s_j}_{|U_i \cap U_j}$. Let $i_0$ such that $x \in U_{i_0}$. We would like to set $s = s_{i_0} \in S = S_x(X)$. But while the other cases are easy to check, how do we check that if $i \in I$ is such that $x \notin U_i$ and $U_i \cap U_{i_0} = \varnothing$, $s_{|U_i} = s_{U_i}$. The problem is that $s_{|U_i} = \phi(s) \in T$ is not necessarily equal to $s_{U_i} \in T$, i.e. $T$ does not necessarily have one element as in the category of groups.

  • 2
    $\begingroup$ If you want to work in an arbitrary category, then you cannot talk about elements of objects, so statements like "$s_i \in S_x(U_i)$" do not make much sense a priori. For presheaves with values in an arbitrary category, the gluing axiom has to be reformulated in such a way that everything can be said just with objects, morphisms and their properties. $\endgroup$ Dec 14, 2016 at 11:53
  • $\begingroup$ Ah ! Thank you. And is it a sheaf under this suitable reformulation ? $\endgroup$ Dec 14, 2016 at 12:00

1 Answer 1


Suppose we have a suitable category $\mathsf{C}$ (namely one where it makes sense to sheafify), we have a canonical isomorphism $\mathsf{C} \cong \mathrm{Sh}(1, \mathsf{C})$, where $1$ denotes the one-point space.

A point $x$ of a space $X$ can be canonically identified with (or even defined as) a map $x: 1 \to X$, and the corresponding direct image functor $x_*: \mathsf{C} \to \mathrm{Sh}(X, \mathsf{C})$ takes an object $C \in \mathsf{C}$ to the skyscraper sheaf with stalk $C$ supported at $x$.

The inverse image $x^{-1}: \mathrm{Sh}(X, \mathsf{C}) \to \mathsf{C}$ is the stalk functor which takes a sheaf $\mathcal{F}$ to its stalk $\mathcal{F}_x$.

Thus, taking stalks at $x$ is left adjoint to the construction of skyscraper sheaves supported at $x$.

If you're wondering how to write down the sheaf condition for presheaves with values in arbitrary category, have a look at section 17.3 in Categories and Sheaves by Kashiwara and Schapira.


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