Definition 4.16.1. We say that a category $\mathcal{I}$ is *connected* if the equivalence relation generated by $x \sim y \Leftrightarrow \mathop{\mathrm{Mor}}\nolimits _\mathcal {I}(x, y) \not= \emptyset $ has exactly one equivalence class.

## 4.16 Connected limits

A (co)limit is called connected if its index category is connected.

Here we follow the convention of Topology, Definition 5.7.1 that connected spaces are nonempty. The following in some vague sense characterizes connected limits.

Lemma 4.16.2. Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let $M : \mathcal{I} \to \mathcal{C}/X$ be a diagram in the category of objects over $X$. If the index category $\mathcal{I}$ is connected and the limit of $M$ exists in $\mathcal{C}/X$, then the limit of the composition $\mathcal{I} \to \mathcal{C}/X \to \mathcal{C}$ exists and is the same.

**Proof.**
Let $L \to X$ be an object representing the limit in $\mathcal{C}/X$. Consider the functor

Let $(\varphi _ i)$ be an element of the set on the right. Since each $M_ i$ comes equipped with a morphism $s_ i : M_ i \to X$ we get morphisms $f_ i = s_ i \circ \varphi _ i : W \to X$. But as $\mathcal{I}$ is connected we see that all $f_ i$ are equal. Since $\mathcal{I}$ is nonempty there is at least one $f_ i$. Hence this common value $W \to X$ defines the structure of an object of $W$ in $\mathcal{C}/X$ and $(\varphi _ i)$ defines an element of $\mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}/X}(W, M_ i)$. Thus we obtain a unique morphism $\phi : W \to L$ such that $\varphi _ i$ is the composition of $\phi $ with $L \to M_ i$ as desired. $\square$

Lemma 4.16.3. Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let $M : \mathcal{I} \to X/\mathcal{C}$ be a diagram in the category of objects under $X$. If the index category $\mathcal{I}$ is connected and the colimit of $M$ exists in $X/\mathcal{C}$, then the colimit of the composition $\mathcal{I} \to X/\mathcal{C} \to \mathcal{C}$ exists and is the same.

**Proof.**
Omitted. Hint: This lemma is dual to Lemma 4.16.2.
$\square$

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