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 $M \to X$ 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 is an element of $\mathop{\mathrm{lim}}\nolimits _ i \mathop{Mor}\nolimits _{\mathcal{C}/X}(W, M_ i)$. Thus we obtain a unique morphism $\phi : W \to M$ such that $\varphi _ i$ is the composition of $\phi $ with $M \to M_ i$ as desired. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)