Concluding that $\mathcal{C}/S$ is complete if $\mathcal{C}$ is complete

Question

I’m currently working through Martin Brandenburg’s Einführung in die Kategorientheorie (Introduction to Category Theory) but got stuck at the second part of the following exercise:

Exercise 6.15. Let $\mathcal{C}$ be a category and let $S \in \mathcal{C}$ be an object. Show that a fibre product over $S$ is the same as a product in the over-$S$-category $\mathcal{C}/S$. Conclude with the help of Exercise 6.9: If $\mathcal{C}$ is complete, then $\mathcal{C}/S$ is also complete.

The previous exercise:

Exercise 6.9. Let $f, g \colon X \to Y$ be morphisms in a category which has fibre products and binary products. Construct the equalizer of $f, g$ as a suitable fibre product $X \times_{(f,g), Y \times Y, \Delta_Y} Y$.

(Translated from German by me, the original text can be found on Google Books.)

Question: How can we conclude that $\mathcal{C}/S$ is complete?

My thoughts so far:

• It follows from the first part of 6.15 that $\mathcal{C}/S$ has arbitrary products (because $\mathcal{C}$ has arbitrary fibre products), so it sufficies to show that $\mathcal{C}/S$ has binary equalizers. By 6.9 this would follow from the existence of binary fibre products.

• To show this I tried to use 6.9 a second time: $\mathcal{C}/S$ has binary fibre products if $(\mathcal{C}/S)/T$ has binary products for every $T \in \mathcal{C}/S$. But I don’t see how this follows from the above; I initially hoped that that $(\mathcal{C}/S)/T$ is equivalent to $\mathcal{C}/T'$ for some $T' \in \mathcal{C}$ or some $\mathcal{C}$-valued functor category, but this doesn’t seem to be the case (from what I can tell).

• One can show the existence of binary fibre products in $\mathcal{C}/S$ (or even arbitrary limits) by hand, similarly to how one constructs products in $\mathcal{C}/S$ from fibre products in $\mathcal{C}$. (This is the approach taken in this question.) But this doesn’t seem to be what the exercise is aiming for.

Answer 1

Actually, if $T:T'\to S$ is any object of $C/S$ then $(C/S)/T$ is isomorphic to $C/T'$. Indeed, an object $f:A\to T'$ of the latter induces a unique object $T\circ f$ of the former, and a map $a:A\to A'$ in $C$ induces a morphism $[f:A\to T']\to [f':A'\to T']$ in the latter if and only if it induces a morphism $T\circ f\to T\circ f'$ in the former.