
## 4.6 Fibre products

Definition 4.6.1. Let $x, y, z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $f\in \mathop{Mor}\nolimits _\mathcal {C}(x, y)$ and $g\in \mathop{Mor}\nolimits _{\mathcal C}(z, y)$. A fibre product of $f$ and $g$ is an object $x \times _ y z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ together with morphisms $p \in \mathop{Mor}\nolimits _{\mathcal C}(x \times _ y z, x)$ and $q \in \mathop{Mor}\nolimits _{\mathcal C}(x \times _ y z, z)$ 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 $\alpha \in \mathop{Mor}\nolimits _{\mathcal C}(w, x)$ and $\beta \in \mathop{Mor}\nolimits _\mathcal {C}(w, z)$ with $f \circ \alpha = g \circ \beta$ there is a unique $\gamma \in \mathop{Mor}\nolimits _{\mathcal C}(w, x \times _ y z)$ 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 $f\in \mathop{Mor}\nolimits _{\mathcal C}(x, y)$ and $g\in \mathop{Mor}\nolimits _{\mathcal C}(z, y)$.

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. Omitted. $\square$

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. $\square$

Comment #153 by on

After 001V, one should add the definition of cartesian square, use without definition for the first time in 0024.

Comment #3413 by Herman Rohrbach on

It might be nice to state a "pasting law for pullbacks" (see e.g. https://ncatlab.org/nlab/show/pasting+law+for+pullbacks). For example, lemma 001Y is a direct consequence of this.

Comment #3472 by on

Maybe, but actually, Lemma 4.6.5 seems totally trivial to me. I guess what I am saying is that the "pasting law" is akin to $1 + 1 = 2$ once you've been sufficiently indoctrinated. I am not trying to be snarky or annoying (just trying to become more like my hero Linus Torvalds...)

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