Lemma 24.2.3. Let $\mathcal{C}$ be a category.

1. the category $\text{SR}(\mathcal{C})$ has coproducts and $F$ commutes with them,

2. the functor $F : \text{SR}(\mathcal{C}) \to \textit{PSh}(\mathcal{C})$ commutes with limits,

3. if $\mathcal{C}$ has fibre products, then $\text{SR}(\mathcal{C})$ has fibre products,

4. if $\mathcal{C}$ has products of pairs, then $\text{SR}(\mathcal{C})$ has products of pairs,

5. if $\mathcal{C}$ has equalizers, so does $\text{SR}(\mathcal{C})$, and

6. if $\mathcal{C}$ has a final object, so does $\text{SR}(\mathcal{C})$.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

1. the category $\text{SR}(\mathcal{C}, X)$ has coproducts and $F$ commutes with them,

2. if $\mathcal{C}$ has fibre products, then $\text{SR}(\mathcal{C}, X)$ has finite limits and $F : \text{SR}(\mathcal{C}, X) \to \textit{PSh}(\mathcal{C})/h_ X$ commutes with them.

Proof. Proof of the results on $\text{SR}(\mathcal{C})$. Proof of (1). The coproduct of $\{ U_ i\} _{i \in I}$ and $\{ V_ j\} _{j \in J}$ is $\{ U_ i\} _{i \in I} \amalg \{ V_ j\} _{j \in J}$, in other words, the family of objects whose index set is $I \amalg J$ and for an element $k \in I \amalg J$ gives $U_ i$ if $k = i \in I$ and gives $V_ j$ if $k = j \in J$. Similarly for coproducts of families of objects. It is clear that $F$ commutes with these.

Proof of (2). For $U$ in $\mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ consider the object $\{ U\}$ of $\text{SR}(\mathcal{C})$. It is clear that $\mathop{Mor}\nolimits _{\text{SR}(\mathcal{C})}(\{ U\} , K)) = F(K)(U)$ for $K \in \mathop{\mathrm{Ob}}\nolimits (\text{SR}(\mathcal{C}))$. Since limits of presheaves are computed at the level of sections (Sites, Section 7.4) we conclude that $F$ commutes with limits.

Proof of (3). Suppose given a morphism $(\alpha , f_ i) : \{ U_ i\} _{i \in I} \to \{ V_ j\} _{j \in J}$ and a morphism $(\beta , g_ k) : \{ W_ k\} _{k \in K} \to \{ V_ j\} _{j \in J}$. The fibred product of these morphisms is given by

$\{ U_ i \times _{f_ i, V_ j, g_ k} W_ k\} _{(i, j, k) \in I \times J \times K \text{ such that } j = \alpha (i) = \beta (k)}$

The fibre products exist if $\mathcal{C}$ has fibre products.

Proof of (4). The product of $\{ U_ i\} _{i \in I}$ and $\{ V_ j\} _{j \in J}$ is $\{ U_ i \times V_ j\} _{i \in I, j \in J}$. The products exist if $\mathcal{C}$ has products.

Proof of (5). The equalizer of two maps $(\alpha , f_ i), (\alpha ', f'_ i) : \{ U_ i\} _{i \in I} \to \{ V_ j\} _{j \in J}$ is

$\{ \text{Eq}(f_ i, f'_ i : U_ i \to V_{\alpha (i)}) \} _{i \in I,\ \alpha (i) = \alpha '(i)}$

The equalizers exist if $\mathcal{C}$ has equalizers.

Proof of (6). If $X$ is a final object of $\mathcal{C}$, then $\{ X\}$ is a final object of $\text{SR}(\mathcal{C})$.

Proof of the statements about $\text{SR}(\mathcal{C}, X)$. These follow from the results above applied to the category $\mathcal{C}/X$ using that $\text{SR}(\mathcal{C}/X) = \text{SR}(\mathcal{C}, X)$ and that $\textit{PSh}(\mathcal{C}/X) = \textit{PSh}(\mathcal{C})/h_ X$ (Sites, Lemma 7.25.4 applied to $\mathcal{C}$ endowed with the chaotic topology). However we also argue directly as follows. It is clear that the coproduct of $\{ U_ i \to X\} _{i \in I}$ and $\{ V_ j \to X\} _{j \in J}$ is $\{ U_ i \to X\} _{i \in I} \amalg \{ V_ j \to X\} _{j \in J}$ and similarly for coproducts of families of families of morphisms with target $X$. The object $\{ X \to X\}$ is a final object of $\text{SR}(\mathcal{C}, X)$. Suppose given a morphism $(\alpha , f_ i) : \{ U_ i \to X\} _{i \in I} \to \{ V_ j \to X\} _{j \in J}$ and a morphism $(\beta , g_ k) : \{ W_ k \to X\} _{k \in K} \to \{ V_ j \to X\} _{j \in J}$. The fibred product of these morphisms is given by

$\{ U_ i \times _{f_ i, V_ j, g_ k} W_ k \to X \} _{(i, j, k) \in I \times J \times K \text{ such that } j = \alpha (i) = \beta (k)}$

The fibre products exist by the assumption that $\mathcal{C}$ has fibre products. Thus $\text{SR}(\mathcal{C}, X)$ has finite limits, see Categories, Lemma 4.18.4. We omit verifying the statements on the functor $F$ in this case. $\square$

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