Lemma 4.6.5. Let $\mathcal{C}$ be a category. Let $f : x \to y$, and $g : y \to z$ be representable. Then $g \circ f : x \to z$ is representable.
Proof. Let $t \in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$ and $ \varphi \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(t,z) $. As $g$ and $f$ are representable, we obtain commutative diagrams
\[ \xymatrix{ y \times _ z t \ar[r]_ q \ar[d]_ p & t \ar[d]^{\varphi } \\ y \ar[r]^{g} & z } \quad \quad \xymatrix{ x \times _ y (y\! \times _ z\! t) \ar[r]_{q'} \ar[d]_{p'} & y \times _ z t \ar[d]^ p \\ x \ar[r]^ f & y } \]
with the universal property of Definition 4.6.1. We claim that $x \times _ z t = x \times _ y (y \times _ z t)$ with morphisms $q \circ q' : x \times _ z t \to t$ and $p' : x \times _ z t \to x$ is a fibre product. First, it follows from the commutativity of the diagrams above that $\varphi \circ q \circ q' = f \circ g \circ p'$. To verify the universal property, let $w \in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$ and suppose $\alpha : w \to x$ and $\beta : w \to y$ are morphisms with $\varphi \circ \beta = f \circ g \circ \alpha $. By definition of the fibre product, there are unique morphisms $\delta $ and $\gamma $ such that
\[ \xymatrix{ w \ar[rrrd]^\beta \ar@{-->}[rrd]_\delta \ar[rrdd]_{f\circ \alpha } & & \\ & & y \times _ z t \ar[d]_ p \ar[r]_ q & t \ar[d]^{\varphi } \\ & & y \ar[r]^{g} & z } \]
and
\[ \xymatrix{ w \ar[rrrd]^\delta \ar@{-->}[rrd]_\gamma \ar[rrdd]_{\alpha } & & \\ & & x \times _ y (y\! \times _ z\! t) \ar[d]_{p'} \ar[r]_{q'} & y \times _ z t \ar[d]^{p} \\ & & x \ar[r]^{f} & y } \]
commute. Then, $\gamma $ makes the diagram
\[ \xymatrix{ w \ar[rrrd]^\beta \ar@{-->}[rrd]_\gamma \ar[rrdd]_{\alpha } & & \\ & & x \times _ z t \ar[d]_{p'} \ar[r]_{q\circ q'} & t \ar[d]^{\varphi } \\ & & x \ar[r]^{g\circ f} & z } \]
commute. To show its uniqueness, let $\gamma '$ verify $ q\circ q'\circ \gamma ' = \beta $ and $ p'\circ \gamma ' = \alpha $. Because $\gamma $ is unique, we just need to prove that $ q'\circ \gamma ' = \delta $ and $ p'\circ \gamma ' = \alpha $ to conclude. We supposed the second equality. For the first one, we also need to use the uniqueness of delta. Notice that $\delta $ is the only morphism verifying $ q\circ \delta = \beta $ and $ p\circ \delta = f\circ \alpha $. We already supposed that $ q\circ (q'\circ \gamma ') = \beta $. Furthermore, by definition of the fibre product, we know that $ f\circ p' = p\circ q' $. Therefore:
\[ p\circ (q'\circ \gamma ') = (p\circ q')\circ \gamma ' = (f\circ p')\circ \gamma ' = f\circ (p'\circ \gamma ') = f\circ \alpha . \]
Then $ q'\circ \gamma ' = \delta $, which concludes the proof. $\square$
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Comment #9862 by Tadahiro Nakajima on
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