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
Comments (1)
Comment #9862 by Tadahiro Nakajima on
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