## 4.31 2-fibre products

In this section we introduce $2$-fibre products. Suppose that $\mathcal{C}$ is a 2-category. We say that a diagram

\[ \xymatrix{ w \ar[r] \ar[d] & y \ar[d] \\ x \ar[r] & z } \]

2-commutes if the two 1-morphisms $w \to y \to z$ and $w \to x \to z$ are 2-isomorphic. In a 2-category it is more natural to ask for 2-commutativity of diagrams than for actually commuting diagrams. (Indeed, some may say that we should not work with strict 2-categories at all, and in a “weak” 2-category the notion of a commutative diagram of 1-morphisms does not even make sense.) Correspondingly the notion of a fibre product has to be adjusted.

Let $\mathcal{C}$ be a $2$-category. Let $x, y, z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $f\in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(x, z)$ and $g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(y, z)$. In order to define the 2-fibre product of $f$ and $g$ we are going to look at 2-commutative diagrams

\[ \xymatrix{ & w \ar[r]_ a \ar[d]_ b & x \ar[d]^{f} \\ & y \ar[r]^{g} & z. } \]

Now in the case of categories, the fibre product is a final object in the category of such diagrams. Correspondingly a 2-fibre product is a final object in a 2-category (see definition below). The *$2$-category of $2$-commutative diagrams over $f$ and $g$* is the $2$-category defined as follows:

Objects are quadruples $(w, a, b, \phi )$ as above where $\phi $ is an invertible 2-morphism $\phi : f \circ a \to g \circ b$,

1-morphisms from $(w', a', b', \phi ')$ to $(w, a, b, \phi )$ are given by $(k : w' \to w, \alpha : a' \to a \circ k, \beta : b' \to b \circ k)$ such that

\[ \xymatrix{ f \circ a' \ar[rr]_{\text{id}_ f \star \alpha } \ar[d]_{\phi '} & & f \circ a \circ k \ar[d]^{\phi \star \text{id}_ k} \\ g \circ b' \ar[rr]^{\text{id}_ g \star \beta } & & g \circ b \circ k } \]

is commutative,

given a second $1$-morphism $(k', \alpha ', \beta ') : (w'', a'', b'', \phi '') \to (w', \alpha ', \beta ', \phi ')$ the composition of $1$-morphisms is given by the rule

\[ (k, \alpha , \beta ) \circ (k', \alpha ', \beta ') = (k \circ k', (\alpha \star \text{id}_{k'}) \circ \alpha ', (\beta \star \text{id}_{k'}) \circ \beta '), \]

a 2-morphism between $1$-morphisms $(k_ i, \alpha _ i, \beta _ i)$, $i = 1, 2$ with the same source and target is given by a 2-morphism $\delta : k_1 \to k_2$ such that

\[ \xymatrix{ a' \ar[rd]_{\alpha _2} \ar[r]_{\alpha _1} & a \circ k_1 \ar[d]^{\text{id}_ a \star \delta } & & b \circ k_1 \ar[d]_{\text{id}_ b \star \delta } & b' \ar[l]^{\beta _1} \ar[ld]^{\beta _2} \\ & a \circ k_2 & & b \circ k_2 & } \]

commute,

vertical composition of $2$-morphisms is given by vertical composition of the morphisms $\delta $ in $\mathcal{C}$, and

horizontal composition of the diagram

\[ \xymatrix{ (w'', a'', b'', \phi '') \rrtwocell ^{(k'_1, \alpha '_1, \beta '_1)}_{(k'_2, \alpha '_2, \beta '_2)}{\delta '} & & (w', a', b', \phi ') \rrtwocell ^{(k_1, \alpha _1, \beta _1)}_{(k_2, \alpha _2, \beta _2)}{\delta } & & (w, a, b, \phi ) } \]

is given by the diagram

\[ \xymatrix@C=12pc{ (w'', a'', b'', \phi '') \rtwocell ^{(k_1 \circ k'_1, (\alpha _1 \star \text{id}_{k'_1}) \circ \alpha '_1, (\beta _1 \star \text{id}_{k'_1}) \circ \beta '_1)}_{(k_2 \circ k'_2, (\alpha _2 \star \text{id}_{k'_2}) \circ \alpha '_2, (\beta _2 \star \text{id}_{k'_2}) \circ \beta '_2)}{\ \ \ \delta \star \delta '} & (w, a, b, \phi ) } \]

Note that if $\mathcal{C}$ is actually a $(2, 1)$-category, the morphisms $\alpha $ and $\beta $ in (2) above are automatically also isomorphisms^{1}. In addition the $2$-category of $2$-commutative diagrams is also a $(2, 1)$-category if $\mathcal{C}$ is a $(2, 1)$-category.

Definition 4.31.1. A *final object* of a $(2, 1)$-category $\mathcal{C}$ is an object $x$ such that

for every $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there is a morphism $y \to x$, and

every two morphisms $y \to x$ are isomorphic by a unique 2-morphism.

Likely, in the more general case of $2$-categories there are different flavours of final objects. We do not want to get into this and hence we only define $2$-fibre products in the $(2, 1)$-case.

Definition 4.31.2. Let $\mathcal{C}$ be a $(2, 1)$-category. Let $x, y, z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $f\in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(x, z)$ and $g\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(y, z)$. A *2-fibre product of $f$ and $g$* is a final object in the category of 2-commutative diagrams described above. If a 2-fibre product exists we will denote it $x \times _ z y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and denote the required morphisms $p\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ z y, x)$ and $q\in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(x \times _ z y, y)$ making the diagram

\[ \xymatrix{ & x \times _ z y \ar[r]^{p} \ar[d]_ q & x \ar[d]^{f} \\ & y \ar[r]^{g} & z } \]

2-commute and we will denote the given invertible 2-morphism exhibiting this by $\psi : f \circ p \to g \circ q$.

Thus the following universal property holds: for any $w\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and morphisms $a \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x)$ and $b \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(w, y)$ with a given 2-isomorphism $\phi : f \circ a \to g\circ b$ there is a $\gamma \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal C}(w, x \times _ z y)$ making the diagram

\[ \xymatrix{ w\ar[rrrd]^ a \ar@{-->}[rrd]_\gamma \ar[rrdd]_ b & & \\ & & x \times _ z y \ar[r]_ p \ar[d]_ q & x \ar[d]^{f} \\ & & y \ar[r]^{g} & z } \]

2-commute such that for suitable choices of $a \to p \circ \gamma $ and $b \to q \circ \gamma $ the diagram

\[ \xymatrix{ f \circ a \ar[r] \ar[d]_\phi & f \circ p \circ \gamma \ar[d]^{\psi \star \text{id}_\gamma } \\ g\circ b \ar[r] & g \circ q \circ \gamma } \]

commutes. Moreover $\gamma $ is unique up to isomorphism. Of course the exact properties are finer than this. All of the cases of 2-fibre products that we will need later on come from the following example of 2-fibre products in the 2-category of categories.

Example 4.31.3. Let $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ be categories. Let $F : \mathcal{A} \to \mathcal{C}$ and $G : \mathcal{B} \to \mathcal{C}$ be functors. We define a category $\mathcal{A} \times _\mathcal {C} \mathcal{B}$ as follows:

an object of $\mathcal{A} \times _\mathcal {C} \mathcal{B}$ is a triple $(A, B, f)$, where $A\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$, $B\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$, and $f : F(A) \to G(B)$ is an isomorphism in $\mathcal{C}$,

a morphism $(A, B, f) \to (A', B', f')$ is given by a pair $(a, b)$, where $a : A \to A'$ is a morphism in $\mathcal{A}$, and $b : B \to B'$ is a morphism in $\mathcal{B}$ such that the diagram

\[ \xymatrix{ F(A) \ar[r]^ f \ar[d]^{F(a)} & G(B) \ar[d]^{G(b)} \\ F(A') \ar[r]^{f'} & G(B') } \]

is commutative.

Moreover, we define functors $p : \mathcal{A} \times _\mathcal {C}\mathcal{B} \to \mathcal{A}$ and $q : \mathcal{A} \times _\mathcal {C}\mathcal{B} \to \mathcal{B}$ by setting

\[ p(A, B, f) = A, \quad q(A, B, f) = B, \]

in other words, these are the forgetful functors. We define a transformation of functors $\psi : F \circ p \to G \circ q$. On the object $\xi = (A, B, f)$ it is given by $\psi _\xi = f : F(p(\xi )) = F(A) \to G(B) = G(q(\xi ))$.

Lemma 4.31.4. In the $(2, 1)$-category of categories $2$-fibre products exist and are given by the construction of Example 4.31.3.

**Proof.**
Let us check the universal property: let $\mathcal{W}$ be a category, let $a : \mathcal{W} \to \mathcal{A}$ and $b : \mathcal{W} \to \mathcal{B}$ be functors, and let $t : F \circ a \to G \circ b$ be an isomorphism of functors.

Consider the functor $\gamma : \mathcal{W} \to \mathcal{A} \times _\mathcal {C}\mathcal{B}$ given by $W \mapsto (a(W), b(W), t_ W)$. (Check this is a functor omitted.) Moreover, consider $\alpha : a \to p \circ \gamma $ and $\beta : b \to q \circ \gamma $ obtained from the identities $p \circ \gamma = a$ and $q \circ \gamma = b$. Then it is clear that $(\gamma , \alpha , \beta )$ is a morphism from $(W, a, b, t)$ to $(\mathcal{A} \times _\mathcal {C} \mathcal{B}, p, q, \psi )$.

Let $(k, \alpha ', \beta ') : (W, a, b, t) \to (\mathcal{A} \times _\mathcal {C} \mathcal{B}, p, q, \psi )$ be a second such morphism. For an object $W$ of $\mathcal{W}$ let us write $k(W) = (a_ k(W), b_ k(W), t_{k, W})$. Hence $p(k(W)) = a_ k(W)$ and so on. The map $\alpha '$ corresponds to functorial maps $\alpha ' : a(W) \to a_ k(W)$. Since we are working in the $(2, 1)$-category of categories, in fact each of the maps $a(W) \to a_ k(W)$ is an isomorphism. We can use these (and their counterparts $b(W) \to b_ k(W)$) to get isomorphisms

\[ \delta _ W : \gamma (W) = (a(W), b(W), t_ W) \longrightarrow (a_ k(W), b_ k(W), t_{k, W}) = k(W). \]

It is straightforward to show that $\delta $ defines a $2$-isomorphism between $\gamma $ and $k$ in the $2$-category of $2$-commutative diagrams as desired.
$\square$

Lemma 4.31.6. Let

\[ \xymatrix{ & \mathcal{Y} \ar[d]_ I \ar[rd]^ K & \\ \mathcal{X} \ar[r]^ H \ar[rd]^ L & \mathcal{Z} \ar[rd]^ M & \mathcal{B} \ar[d]^ G \\ & \mathcal{A} \ar[r]^ F & \mathcal{C} } \]

be a $2$-commutative diagram of categories. A choice of isomorphisms $\alpha : G \circ K \to M \circ I$ and $\beta : M \circ H \to F \circ L$ determines a morphism

\[ \mathcal{X} \times _\mathcal {Z} \mathcal{Y} \longrightarrow \mathcal{A} \times _\mathcal {C} \mathcal{B} \]

of $2$-fibre products associated to this situation.

**Proof.**
Just use the functor

\[ (X, Y, \phi ) \longmapsto (L(X), K(Y), \alpha ^{-1}_ Y \circ M(\phi ) \circ \beta ^{-1}_ X) \]

on objects and

\[ (a, b) \longmapsto (L(a), K(b)) \]

on morphisms.
$\square$

Lemma 4.31.7. Assumptions as in Lemma 4.31.6.

If $K$ and $L$ are faithful then the morphism $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{A} \times _\mathcal {C} \mathcal{B}$ is faithful.

If $K$ and $L$ are fully faithful and $M$ is faithful then the morphism $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{A} \times _\mathcal {C} \mathcal{B}$ is fully faithful.

If $K$ and $L$ are equivalences and $M$ is fully faithful then the morphism $\mathcal{X} \times _\mathcal {Z} \mathcal{Y} \to \mathcal{A} \times _\mathcal {C} \mathcal{B}$ is an equivalence.

**Proof.**
Let $(X, Y, \phi )$ and $(X', Y', \phi ')$ be objects of $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$. Set $Z = H(X)$ and identify it with $I(Y)$ via $\phi $. Also, identify $M(Z)$ with $F(L(X))$ via $\alpha _ X$ and identify $M(Z)$ with $G(K(Y))$ via $\beta _ Y$. Similarly for $Z' = H(X')$ and $M(Z')$. The map on morphisms is the map

\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits _\mathcal {X}(X, X') \times _{\mathop{\mathrm{Mor}}\nolimits _\mathcal {Z}(Z, Z')} \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(Y, Y') \ar[d] \\ \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(L(X), L(X')) \times _{\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(M(Z), M(Z'))} \mathop{\mathrm{Mor}}\nolimits _\mathcal {B}(K(Y), K(Y')) } \]

Hence parts (1) and (2) follow. Moreover, if $K$ and $L$ are equivalences and $M$ is fully faithful, then any object $(A, B, \phi )$ is in the essential image for the following reasons: Pick $X$, $Y$ such that $L(X) \cong A$ and $K(Y) \cong B$. Then the fully faithfulness of $M$ guarantees that we can find an isomorphism $H(X) \cong I(Y)$. Some details omitted.
$\square$

Lemma 4.31.8. Let

\[ \xymatrix{ \mathcal{A} \ar[rd] & & \mathcal{C} \ar[ld] \ar[rd] & & \mathcal{E} \ar[ld] \\ & \mathcal{B} & & \mathcal{D} } \]

be a diagram of categories and functors. Then there is a canonical isomorphism

\[ (\mathcal{A} \times _\mathcal {B} \mathcal{C}) \times _\mathcal {D} \mathcal{E} \cong \mathcal{A} \times _\mathcal {B} (\mathcal{C} \times _\mathcal {D} \mathcal{E}) \]

of categories.

**Proof.**
Just use the functor

\[ ((A, C, \phi ), E, \psi ) \longmapsto (A, (C, E, \psi ), \phi ) \]

if you know what I mean.
$\square$

Henceforth we do not write the parentheses when dealing with fibre products of more than 2 categories.

Lemma 4.31.9. Let

\[ \xymatrix{ \mathcal{A} \ar[rd] & & \mathcal{C} \ar[ld] \ar[rd] & & \mathcal{E} \ar[ld] \\ & \mathcal{B} \ar[rd]_ F & & \mathcal{D} \ar[ld]^ G \\ & & \mathcal{F} & } \]

be a commutative diagram of categories and functors. Then there is a canonical functor

\[ \text{pr}_{02} : \mathcal{A} \times _\mathcal {B} \mathcal{C} \times _\mathcal {D} \mathcal{E} \longrightarrow \mathcal{A} \times _\mathcal {F} \mathcal{E} \]

of categories.

**Proof.**
If we write $\mathcal{A} \times _\mathcal {B} \mathcal{C} \times _\mathcal {D} \mathcal{E}$ as $(\mathcal{A} \times _\mathcal {B} \mathcal{C}) \times _\mathcal {D} \mathcal{E}$ then we can just use the functor

\[ ((A, C, \phi ), E, \psi ) \longmapsto (A, E, G(\psi ) \circ F(\phi )) \]

if you know what I mean.
$\square$

Lemma 4.31.10. Let

\[ \mathcal{A} \to \mathcal{B} \leftarrow \mathcal{C} \leftarrow \mathcal{D} \]

be a diagram of categories and functors. Then there is a canonical isomorphism

\[ \mathcal{A} \times _\mathcal {B} \mathcal{C} \times _\mathcal {C} \mathcal{D} \cong \mathcal{A} \times _\mathcal {B} \mathcal{D} \]

of categories.

**Proof.**
Omitted.
$\square$

We claim that this means you can work with these $2$-fibre products just like with ordinary fibre products. Here are some further lemmas that actually come up later.

Lemma 4.31.11. Let

\[ \xymatrix{ \mathcal{C}_3 \ar[r] \ar[d] & \mathcal{S} \ar[d]^\Delta \\ \mathcal{C}_1 \times \mathcal{C}_2 \ar[r]^{G_1 \times G_2} & \mathcal{S} \times \mathcal{S} } \]

be a $2$-fibre product of categories. Then there is a canonical isomorphism $\mathcal{C}_3 \cong \mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$.

**Proof.**
We may assume that $\mathcal{C}_3$ is the category $(\mathcal{C}_1 \times \mathcal{C}_2)\times _{\mathcal{S} \times \mathcal{S}} \mathcal{S}$ constructed in Example 4.31.3. Hence an object is a triple $((X_1, X_2), S, \phi )$ where $\phi = (\phi _1, \phi _2) : (G_1(X_1), G_2(X_2)) \to (S, S)$ is an isomorphism. Thus we can associate to this the triple $(X_1, X_2, \phi _2^{-1} \circ \phi _1)$. Conversely, if $(X_1, X_2, \psi )$ is an object of $\mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$, then we can associate to this the triple $((X_1, X_2), G_2(X_2), (\psi , \text{id}_{G_2(X_2)}))$. We claim these constructions given mutually inverse functors. We omit describing how to deal with morphisms and showing they are mutually inverse.
$\square$

Lemma 4.31.12. Let

\[ \xymatrix{ \mathcal{C}' \ar[r] \ar[d] & \mathcal{S} \ar[d]^\Delta \\ \mathcal{C} \ar[r]^{G_1 \times G_2} & \mathcal{S} \times \mathcal{S} } \]

be a $2$-fibre product of categories. Then there is a canonical isomorphism

\[ \mathcal{C}' \cong (\mathcal{C} \times _{G_1, \mathcal{S}, G_2} \mathcal{C}) \times _{(p, q), \mathcal{C} \times \mathcal{C}, \Delta } \mathcal{C}. \]

**Proof.**
An object of the right hand side is given by $((C_1, C_2, \phi ), C_3, \psi )$ where $\phi : G_1(C_1) \to G_2(C_2)$ is an isomorphism and $\psi = (\psi _1, \psi _2) : (C_1, C_2) \to (C_3, C_3)$ is an isomorphism. Hence we can associate to this the triple $(C_3, G_1(C_1), (G_1(\psi _1^{-1}), \phi ^{-1} \circ G_2(\psi _2^{-1})))$ which is an object of $\mathcal{C}'$. Details omitted.
$\square$

Lemma 4.31.13. Let $\mathcal{A} \to \mathcal{C}$, $\mathcal{B} \to \mathcal{C}$ and $\mathcal{C} \to \mathcal{D}$ be functors between categories. Then the diagram

\[ \xymatrix{ \mathcal{A} \times _\mathcal {C} \mathcal{B} \ar[d] \ar[r] & \mathcal{A} \times _\mathcal {D} \mathcal{B} \ar[d] \\ \mathcal{C} \ar[r]^-{\Delta _{\mathcal{C}/\mathcal{D}}} \ar[r] & \mathcal{C} \times _\mathcal {D} \mathcal{C} } \]

is a $2$-fibre product diagram.

**Proof.**
Omitted.
$\square$

Lemma 4.31.14. Let

\[ \xymatrix{ \mathcal{U} \ar[d] \ar[r] & \mathcal{V} \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} } \]

be a $2$-fibre product of categories. Then the diagram

\[ \xymatrix{ \mathcal{U} \ar[d] \ar[r] & \mathcal{U} \times _\mathcal {V} \mathcal{U} \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

is $2$-cartesian.

**Proof.**
This is a purely $2$-category theoretic statement, valid in any $(2, 1)$-category with $2$-fibre products. Explicitly, it follows from the following chain of equivalences:

\begin{align*} \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Y} \mathcal{X})} (\mathcal{U} \times _\mathcal {V} \mathcal{U}) & = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Y} \mathcal{X})} ((\mathcal{X} \times _\mathcal {Y} \mathcal{V}) \times _\mathcal {V} (\mathcal{X} \times _\mathcal {Y} \mathcal{V})) \\ & = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Y} \mathcal{X})} (\mathcal{X} \times _\mathcal {Y} \mathcal{X} \times _\mathcal {Y} \mathcal{V}) \\ & = \mathcal{X} \times _\mathcal {Y} \mathcal{V} = \mathcal{U} \end{align*}

see Lemmas 4.31.8 and 4.31.10.
$\square$

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