Lemma 7.34.1. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Let $v : \mathcal{D} \to \textit{Sets}$ be a functor and set $w = v \circ u$. Denote $q$, resp., $p$ the stalk functor (7.32.1.1) associated to $v$, resp. $w$. Then $(u_ p\mathcal{F})_ q = \mathcal{F}_ p$ functorially in the presheaf $\mathcal{F}$ on $\mathcal{C}$.

## 7.34 Points and morphisms of topoi

In this section we make a few remarks about points and morphisms of topoi.

**Proof.**
This is a simple categorical fact. We have

The first equality holds by the definition of $u_ p$ and the definition of the stalk functor. Observe that $y \in v(V)$. In the second equality we simply combine colimits. To see the third equality we apply Categories, Lemma 4.17.5 to the functor $F$ of diagram categories defined by the rule

This makes sense because $w(U) = v(u(U))$. Let us check the hypotheses of Categories, Lemma 4.17.5. Observe that $F$ has a right inverse, namely $(U, x) \mapsto (u(U), x, U, \text{id} : u(U) \to u(U))$. Again this makes sense because $x \in w(U) = v(u(U))$. On the other hand, there is always a morphism

in the fibre category over $(U, x)$ which shows the fibre categories are connected. The fourth and final equality is clear. $\square$

Lemma 7.34.2. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by a continuous functor $u : \mathcal{C} \to \mathcal{D}$. Let $q$ be a point of $\mathcal{D}$ given by the functor $v : \mathcal{D} \to \textit{Sets}$, see Definition 7.32.2. Then the functor $v \circ u : \mathcal{C} \to \textit{Sets}$ defines a point $p$ of $\mathcal{C}$ and moreover there is a canonical identification

for any sheaf $\mathcal{F}$ on $\mathcal{C}$.

**First proof Lemma 7.34.2.**
Note that since $u$ is continuous and since $v$ defines a point, it is immediate that $v \circ u$ satisfies conditions (1) and (2) of Definition 7.32.2. Let us prove the displayed equality. Let $\mathcal{F}$ be a sheaf on $\mathcal{C}$. Then

The first equality since $f^{-1} = u_ s$, the second equality by Lemma 7.32.5, and the third by Lemma 7.34.1. Hence now we see that $p$ also satisfies condition (3) of Definition 7.32.2 because it is a composition of exact functors. This finishes the proof. $\square$

**Second proof Lemma 7.34.2.**
By Lemma 7.32.8 we may factor $(q_*, q^{-1})$ as

where the second morphism of topoi comes from a morphism of sites $h : \mathcal{S} \to \mathcal{D}$ induced by the functor $v : \mathcal{D} \to \mathcal{S}$ (which makes sense as $\mathcal{S} \subset \textit{Sets}$ is a full subcategory containing every object in the image of $v$). By Lemma 7.14.4 the composition $v \circ u : \mathcal{C} \to \mathcal{S}$ defines a morphism of sites $g : \mathcal{S} \to \mathcal{C}$. In particular, the functor $v \circ u : \mathcal{C} \to \mathcal{S}$ is continuous which by the definition of the coverings in $\mathcal{S}$, see Remark 7.15.3, means that $v \circ u$ satisfies conditions (1) and (2) of Definition 7.32.2. On the other hand, we see that

by the construction of $i$ in Remark 7.15.3. Note that this is the same as the formula for which is equal to $(v \circ u)^ pE$, see Equation (7.32.3.1). By Lemma 7.32.5 the functor $g_*i_* = (v \circ u)^ p = (v \circ u)^ s$ is right adjoint to the stalk functor $\mathcal{F} \mapsto \mathcal{F}_ q$. Hence we see that the stalk functor $p^{-1}$ is canonically isomorphic to $i^{-1} \circ g^{-1}$. Hence it is exact and we conclude that $p$ is a point. Finally, as we have $g = f \circ h$ by construction we see that $p^{-1} = i^{-1} \circ h^{-1} \circ f^{-1} = q^{-1} \circ f^{-1}$, i.e., we have the displayed formula of the lemma. $\square$

Lemma 7.34.3. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ be a morphism of topoi. Let $q : \mathop{\mathit{Sh}}\nolimits (pt) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be a point. Then $p = f \circ q$ is a point of the topos $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ and we have a canonical identification

for any sheaf $\mathcal{F}$ on $\mathcal{C}$.

**Proof.**
This is immediate from the definitions and the fact that we can compose morphisms of topoi.
$\square$

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