4.6 Fibre products
Definition 4.6.1. Let x, y, z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), f\in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(x, y) and g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(z, y). A fibre product of f and g is an object x \times _ y z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) together with morphisms p \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ y z, x) and q \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ y z, z) making the diagram
\xymatrix{ x \times _ y z \ar[r]_ q \ar[d]_ p & z \ar[d]^ g \\ x \ar[r]^ f & y }
commute, and such that the following universal property holds: for any w\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) and morphisms \alpha \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x) and \beta \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(w, z) with f \circ \alpha = g \circ \beta there is a unique \gamma \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x \times _ y z) making the diagram
\xymatrix{ w \ar[rrrd]^\beta \ar@{-->}[rrd]_\gamma \ar[rrdd]_\alpha & & \\ & & x \times _ y z \ar[d]^ p \ar[r]_ q & z \ar[d]^ g \\ & & x \ar[r]^ f & y }
commute.
If a fibre product exists it is unique up to unique isomorphism. This follows from the Yoneda lemma as the definition requires x \times _ y z to be an object of \mathcal{C} such that
h_{x \times _ y z}(w) = h_ x(w) \times _{h_ y(w)} h_ z(w)
functorially in w. In other words the fibre product x \times _ y z is an object representing the functor w \mapsto h_ x(w) \times _{h_ y(w)} h_ z(w).
Definition 4.6.2. We say a commutative diagram
\xymatrix{ w \ar[r] \ar[d] & z \ar[d] \\ x \ar[r] & y }
in a category is cartesian if w and the morphisms w \to x and w \to z form a fibre product of the morphisms x \to y and z \to y.
Definition 4.6.3. We say the category \mathcal{C} has fibre products if the fibre product exists for any f\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x, y) and g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(z, y).
Definition 4.6.4. A morphism f : x \to y of a category \mathcal{C} is said to be representable if for every morphism z \to y in \mathcal{C} the fibre product x \times _ y z exists.
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
Lemma 4.6.6. Let \mathcal{C} be a category. Let f : x \to y be representable. Let y' \to y be a morphism of \mathcal{C}. Then the morphism x' := x \times _ y y' \to y' is representable also.
Proof.
Let z \to y' be a morphism. The fibre product x' \times _{y'} z is supposed to represent the functor
\begin{eqnarray*} w & \mapsto & h_{x'}(w)\times _{h_{y'}(w)} h_ z(w) \\ & = & (h_ x(w) \times _{h_ y(w)} h_{y'}(w)) \times _{h_{y'}(w)} h_ z(w) \\ & = & h_ x(w) \times _{h_ y(w)} h_ z(w) \end{eqnarray*}
which is representable by assumption.
\square
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