# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

### 4.2.2 Joins of Categories

Our next goal is to characterize the slice categories of Construction 4.2.1.7 by a universal mapping property.

Definition 4.2.2.1 (Joins of Categories). Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be categories. We define a category $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ as follows:

• The set of objects $\operatorname{Ob}( \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})$ is the disjoint union of $\operatorname{Ob}(\operatorname{\mathcal{C}})$ with $\operatorname{Ob}(\operatorname{\mathcal{D}})$.

• Given a pair of objects $X,Y \in \operatorname{Ob}(\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})$, we have

$\operatorname{Hom}_{\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}}(X,Y) = \begin{cases} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) & \textnormal{if } X,Y \in \operatorname{Ob}(\operatorname{\mathcal{C}}) \\ \operatorname{Hom}_{\operatorname{\mathcal{D}}}(X,Y) & \textnormal{if } X,Y \in \operatorname{Ob}(\operatorname{\mathcal{C}}') \\ \ast & \textnormal{if } X \in \operatorname{Ob}(\operatorname{\mathcal{C}}), Y \in \operatorname{Ob}(\operatorname{\mathcal{D}}) \\ \emptyset & \textnormal{if } X \in \operatorname{Ob}(\operatorname{\mathcal{C}}), Y \in \operatorname{Ob}(\operatorname{\mathcal{D}}). \end{cases}$
• Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms in $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$. If $X,Y,Z \in \operatorname{Ob}(\operatorname{\mathcal{C}})$, then $g \circ f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}}(X,Z)$ is given by the composition of morphisms in $\operatorname{\mathcal{C}}$. If $X,Y,Z \in \operatorname{Ob}(\operatorname{\mathcal{D}})$, then $g \circ f$ is given by composition of morphisms in $\operatorname{\mathcal{D}}$. Otherwise, we let $g \circ f$ denote the unique morphism from $X$ to $Z$ (note that in this case, we necessarily have $X \in \operatorname{Ob}(\operatorname{\mathcal{C}})$ and $Z \in \operatorname{Ob}(\operatorname{\mathcal{D}})$).

We will refer to $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ as the join of $\operatorname{\mathcal{C}}$ with $\operatorname{\mathcal{D}}$.

Remark 4.2.2.2. In the situation of Definition 4.2.2.1, we will generally abuse notation by identifying $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ with full subcategories of the join $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$.

Remark 4.2.2.3. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ and $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}'$ be functors. Then $F$ and $G$ induce a functor

$(F \star G): \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}' \star \operatorname{\mathcal{D}}',$

which is uniquely determined by the requirement that it coincides with $F$ on the full subcategory $\operatorname{\mathcal{C}}\subseteq \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ and with $G$ on the full subcategory $\operatorname{\mathcal{D}}\subseteq \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$. We can therefore regard the join construction as a functor

$\star : \operatorname{Cat}\times \operatorname{Cat}\rightarrow \operatorname{Cat}\quad \quad (\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \mapsto \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}},$

where $\operatorname{Cat}$ denotes the category of (small) categories.

Example 4.2.2.4. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be categories. If $\operatorname{\mathcal{D}}$ is empty, then the inclusion map $\operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ is an isomorphism of categories.

Example 4.2.2.5 (Cones). Let $[0]$ denote the category having a single object and a single morphism, and let $\operatorname{\mathcal{C}}$ be an arbitrary category. We let $\operatorname{\mathcal{C}}^{\triangleleft }$ denote the join $[0] \star \operatorname{\mathcal{C}}$, and $\operatorname{\mathcal{C}}^{\triangleright }$ the join $\operatorname{\mathcal{C}}\star [0]$. We refer to $\operatorname{\mathcal{C}}^{\triangleleft }$ as the left cone of $\operatorname{\mathcal{C}}$, and to $\operatorname{\mathcal{C}}^{\triangleright }$ as the right cone on $\operatorname{\mathcal{C}}$.

More informally, we can describe the left cone $\operatorname{\mathcal{C}}^{\triangleleft }$ as the category obtained from $\operatorname{\mathcal{C}}$ by adjoining a new object $X_0$ satisfying

$\operatorname{Hom}_{\operatorname{\mathcal{C}}^{\triangleleft }}( X_0, Y) = \ast \quad \quad \operatorname{Hom}_{\operatorname{\mathcal{C}}^{\triangleleft }}( X_0, X_0 ) = \ast \quad \quad \operatorname{Hom}_{\operatorname{\mathcal{C}}^{\triangleleft }}( Y, X_0 ) = \emptyset$

for $Y \in \operatorname{\mathcal{C}}$. Note that $X_0$ is an initial object of the category $\operatorname{\mathcal{C}}^{\triangleleft }$, which we will refer to as the cone point of $\operatorname{\mathcal{C}}^{\triangleleft }$. Similarly, the right cone $\operatorname{\mathcal{C}}^{\triangleright }$ is obtained from $\operatorname{\mathcal{C}}$ by adjoining a new object which we refer to as the cone point of $\operatorname{\mathcal{C}}^{\triangleright }$ (and which is a final object of $\operatorname{\mathcal{C}}^{\triangleright }$).

Remark 4.2.2.6. Let $\operatorname{\mathcal{C}}$, $\operatorname{\mathcal{D}}$, and $\operatorname{\mathcal{E}}$ be categories. Then there is a canonical isomorphism of iterated joins

$\alpha : \operatorname{\mathcal{C}}\star (\operatorname{\mathcal{D}}\star \operatorname{\mathcal{E}}) \simeq (\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}) \star \operatorname{\mathcal{E}},$

characterized by the requirement that it restricts to the identity on $\operatorname{\mathcal{C}}$, $\operatorname{\mathcal{D}}$, and $\operatorname{\mathcal{E}}$ (which we can regard as full subcategories of both $\operatorname{\mathcal{C}}\star (\operatorname{\mathcal{D}}\star \operatorname{\mathcal{E}})$ and $(\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}) \star \operatorname{\mathcal{E}}$, by means of Remark 4.2.2.2).

Remark 4.2.2.7. Let $\operatorname{Cat}$ denote the category of (small) categories. Then $\operatorname{Cat}$ admits a monoidal structure, where the tensor product is given by the join functor

$\star : \operatorname{Cat}\times \operatorname{Cat}\rightarrow \operatorname{Cat}\quad \quad (\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \mapsto \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$

of Remark 4.2.2.3, and the associativity constraints are the isomorphisms of Remark 4.2.2.6. The unit for this monoidal structure is the empty category $\emptyset \in \operatorname{Cat}$ (Example 4.2.2.4).

Warning 4.2.2.8. The join operation of Definition 4.2.2.1 is not commutative. For example, if $\operatorname{\mathcal{C}}$ is a category, then the left cone $\operatorname{\mathcal{C}}^{\triangleleft }$ need not be isomorphic (or even equivalent) to the right cone $\operatorname{\mathcal{C}}^{\triangleright }$. However, we do have canonical isomorphisms

$(\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})^{\operatorname{op}} \simeq \operatorname{\mathcal{D}}^{\operatorname{op}} \star \operatorname{\mathcal{C}}^{\operatorname{op}},$

depending functorially on $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$.

We now relate the join construction of Definition 4.2.2.1 with the slice categories of Construction 4.2.1.7. We begin with a simple observation.

Lemma 4.2.2.9. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be categories, and let $\iota _{\operatorname{\mathcal{C}}}: \operatorname{\mathcal{C}}\hookrightarrow \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ and $\iota _{\operatorname{\mathcal{D}}}: \operatorname{\mathcal{D}}\hookrightarrow \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ denote the inclusion maps. Then:

$(1)$

The inclusion functor $\iota _{\operatorname{\mathcal{C}}}$ factors uniquely as a composition

$\operatorname{\mathcal{C}}\xrightarrow { \overline{\iota }_{\operatorname{\mathcal{C}}} } (\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})_{/ \iota _{\operatorname{\mathcal{D}}} } \rightarrow \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}.$
$(2)$

The inclusion functor $\iota _{\operatorname{\mathcal{D}}}$ factors uniquely as a composition

$\operatorname{\mathcal{D}}\xrightarrow { \overline{\iota }_{\operatorname{\mathcal{D}}} } (\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})_{ \iota _{\operatorname{\mathcal{C}}} / } \rightarrow \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}.$

Proof. Let $\pi _{\operatorname{\mathcal{C}}}: \operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ and $\pi _{\operatorname{\mathcal{D}}}: \operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}$ denote the projection maps. Using Remark 4.2.1.10, we see that both $(1)$ and $(2)$ are equivalent to the assertion that there is a unique natural transformation $u$ from $\iota _{\operatorname{\mathcal{C}}} \circ \pi _{\operatorname{\mathcal{C}}}$ to $\iota _{\operatorname{\mathcal{D}}} \circ \pi _{\operatorname{\mathcal{D}}}$ (as functors from the product category $\operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}$ to the join category $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$). Concretely, this natural transformation carries each object $(C,D) \in \operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}$ to the unique element of $\operatorname{Hom}_{\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}}(C, D)$. $\square$

Proposition 4.2.2.10. Let $\operatorname{\mathcal{C}}$ be a category and let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be a functor between categories. For every functor $U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ extending $G$, let $\overline{F}(U)$ denote the composite functor

$\operatorname{\mathcal{C}}\xrightarrow { \overline{\iota }_{\operatorname{\mathcal{C}}} } ( \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})_{ / \iota _{\operatorname{\mathcal{D}}} } \xrightarrow {U } \operatorname{\mathcal{E}}_{ / (U \circ \iota _{\operatorname{\mathcal{D}}} ) } = \operatorname{\mathcal{E}}_{/G}.$

Then the construction $U \mapsto \overline{F}(U)$ induces a bijection

$\{ \textnormal{Functors U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}} satisfying U|_{\operatorname{\mathcal{D}}} = G} \} \rightarrow \{ \textnormal{Functors \overline{F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}_{/G}} \} .$

Example 4.2.2.11. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be a functor of categories. Applying Proposition 4.2.2.10 in the case $\operatorname{\mathcal{C}}= [0]$, we see that objects of the slice category $\operatorname{\mathcal{E}}_{/G}$ can be identified with functors $U: \operatorname{\mathcal{D}}^{\triangleleft } \rightarrow \operatorname{\mathcal{E}}$ satisfying $U|_{\operatorname{\mathcal{D}}} = G$.

Example 4.2.2.12. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{E}}$ be categories and let $S$ be an object of $\operatorname{\mathcal{E}}$. Applying Proposition 4.2.2.10 in the case $\operatorname{\mathcal{D}}= [0]$, we see that functors from $\operatorname{\mathcal{C}}$ to the slice category $\operatorname{\mathcal{E}}_{/S}$ can be identified with functors $U: \operatorname{\mathcal{C}}^{\triangleright } \rightarrow \operatorname{\mathcal{E}}$ which carry the cone point of $\operatorname{\mathcal{C}}^{\triangleright }$ to the object $S$.

In the situation of Proposition 4.2.2.10, we can use Remark 4.2.1.10 to identify functors $\overline{F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}_{/G}$ with ordered pairs $(F, v)$, where $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ is a functor (given by the composition of $\overline{F}$ with the forgetful functor $\operatorname{\mathcal{E}}_{/G} \rightarrow \operatorname{\mathcal{E}}$) and $v$ is a natural transformation from $F \circ \pi _{\operatorname{\mathcal{C}}}$ to $G \circ \pi _{\operatorname{\mathcal{D}}}$ (regarded as functors from $\operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}$ to $\operatorname{\mathcal{E}}$). Note that, in the case where $\overline{F} = \overline{F}(U)$ is obtained from a functor $U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$, we have $F = U|_{\operatorname{\mathcal{C}}}$. We can therefore reformulate Proposition 4.2.2.10 in a more symmetric fashion:

Proposition 4.2.2.13. Let $\operatorname{\mathcal{C}}$, $\operatorname{\mathcal{D}}$, and $\operatorname{\mathcal{E}}$ be categories, and suppose we are given functors $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ and $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$. Let $u: \iota _{\operatorname{\mathcal{C}}} \circ \pi _{\operatorname{\mathcal{C}}} \rightarrow \iota _{\operatorname{\mathcal{D}}} \circ \pi _{\operatorname{\mathcal{D}}}$ be the natural transformation appearing in the proof of Lemma 4.2.2.9. Then evaluation on $u$ induces a bijection

$\xymatrix { \{ \textnormal{Functors U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}} with U|_{\operatorname{\mathcal{C}}} = F and U|_{\operatorname{\mathcal{D}}} = G} \} \ar [d] \\ \{ \textnormal{Natural transformations from F \circ \pi _{\operatorname{\mathcal{C}}} to G \circ \pi _{\operatorname{\mathcal{D}}} } \} }.$

Proof. Let $v$ be a natural transformation from $F \circ \pi _{\operatorname{\mathcal{C}}}$ to $G \circ \pi _{\operatorname{\mathcal{D}}}$, carrying each object $(C,D) \in \operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}$ to a morphism $v_{C,D}: F(C) \rightarrow G(D)$ in the category $\operatorname{\mathcal{E}}$. We wish to show that there is a unique functor $U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ satisfying $U|_{\operatorname{\mathcal{C}}} = F$, $H|_{\operatorname{\mathcal{D}}} = G$, and $U( u_{C,D} ) = v_{C,D}$ for $(C,D) \in \operatorname{\mathcal{C}}\times \operatorname{\mathcal{D}}$. These requirements uniquely determine the value of $U$ on all objects and morphisms of the category $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$. To complete the proof, it will suffice to show that $U$ is compatible with composition: that is, for every pair of morphisms $s: X \rightarrow Y$ and $t: Y \rightarrow Z$ in $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$, we have $U(t \circ s) = U(t) \circ U(s)$. We consider four cases:

• If $X$, $Y$, and $Z$ belong to $\operatorname{\mathcal{C}}$, then we have $U(t \circ s) = F(t \circ s) = F(t) \circ F(s) = U(t) \circ U(s)$.

• If $X$ and $Y$ belong to $\operatorname{\mathcal{C}}$ and $Z$ belongs to $\operatorname{\mathcal{D}}$, then we have $U(t \circ s) = v_{X,Z} = v_{Y,Z} \circ F(s) = U(t) \circ U(s)$, where the second equality follows from the naturality of $v$ in the first variable.

• If $Y$ and $Z$ belong to $\operatorname{\mathcal{D}}$ and $X$ belongs to $\operatorname{\mathcal{C}}$, then we have $U(t \circ s) = v_{X,Z} = G(t) \circ v_{X,Y} = U(t) \circ U(s)$, where the second equality follows from the naturality of $v$ in the second variable.

• If $X$, $Y$, and $Z$ belong to $\operatorname{\mathcal{D}}$, then we have $U(t \circ s) = G(t \circ s) = G(t) \circ G(s) = U(t) \circ U(s)$.

$\square$

Remark 4.2.2.14. Stated more informally, Proposition 4.2.2.13 asserts that the join $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$ is universal among categories $\operatorname{\mathcal{E}}$ which are equipped with a pair of functors $\operatorname{\mathcal{C}}\xrightarrow {F} \operatorname{\mathcal{E}}\xleftarrow {G} \operatorname{\mathcal{D}}$ and a natural transformation $v: (F \circ \pi _{\operatorname{\mathcal{C}}}) \rightarrow (G \circ \pi _{\operatorname{\mathcal{D}}})$. More precisely, there is a pushout square

$\xymatrix@R =50pt@C=50pt{ (\operatorname{\mathcal{C}}\times \{ 0 \} \times \operatorname{\mathcal{D}}) \coprod ( \operatorname{\mathcal{C}}\times \{ 1\} \times \operatorname{\mathcal{D}}) \ar [r] \ar [d] & \operatorname{\mathcal{C}}\times [1] \times \operatorname{\mathcal{D}}\ar [d] \\ ( \operatorname{\mathcal{C}}\times \{ 0\} ) \coprod ( \{ 1\} \times \operatorname{\mathcal{D}}) \ar [r] & \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}}$

in the (ordinary) category $\operatorname{Cat}$, where the right vertical map encodes the natural transformation $u: \iota _{\operatorname{\mathcal{C}}} \circ \pi _{\operatorname{\mathcal{C}}} \rightarrow \iota _{\operatorname{\mathcal{D}}} \circ \pi _{\operatorname{\mathcal{D}}}$ appearing in the proof of Lemma 4.2.2.9.

Example 4.2.2.15 (The Universal Property of a Cone). Let $\operatorname{\mathcal{C}}$ be a category. Applying Remark 4.2.2.14 in the special case $\operatorname{\mathcal{D}}= [0]$, we obtain a pushout diagram of categories

$\xymatrix { \operatorname{\mathcal{C}}\times \{ 1\} \ar [r] \ar [d] & \operatorname{\mathcal{C}}\times {[1]} \ar [d] \\ {[0]} \ar [r] & \operatorname{\mathcal{C}}^{\triangleright }, }$

where the bottom horizontal map carries the unique object of $[0]$ to the cone point of $\operatorname{\mathcal{C}}^{\triangleright }$. This is essentially a reformulation of Examples 4.2.2.11 and 4.2.2.12. Stated more informally, the right cone $\operatorname{\mathcal{C}}^{\triangleright }$ is obtained from the product $[1] \times \operatorname{\mathcal{C}}$ by “collapsing” the full subcategory $\{ 1\} \times \operatorname{\mathcal{C}}$ to the cone point. Similarly, the left cone of a category $\operatorname{\mathcal{D}}$ is characterized by the existence of a pushout diagram

$\xymatrix { \{ 0\} \times \operatorname{\mathcal{D}}\ar [r] \ar [d] & [1] \times \operatorname{\mathcal{D}}\ar [d] \\ {[0]} \ar [r] & \operatorname{\mathcal{D}}^{\triangleleft }. }$

For completeness, we record the dual of Proposition 4.2.2.10, which supplies a universal property of coslice categories (and is also a reformulation of Proposition 4.2.2.13).

Corollary 4.2.2.16. Let $\operatorname{\mathcal{D}}$ be a category and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ be a functor between categories. For every functor $U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ extending $F$, let $\overline{G}(U)$ denote the composite functor

$\operatorname{\mathcal{D}}\xrightarrow { \overline{\iota }_{\operatorname{\mathcal{D}}} } ( \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}})_{ \iota _{\operatorname{\mathcal{C}}} / } \xrightarrow {U } \operatorname{\mathcal{E}}_{ (U \circ \iota _{\operatorname{\mathcal{C}}} )/ } = \operatorname{\mathcal{E}}_{F/}.$

Then the construction $U \mapsto \overline{G}(U)$ induces a bijection

$\{ \textnormal{Functors U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}} satisfying U|_{\operatorname{\mathcal{C}}} = F} \} \rightarrow \{ \textnormal{Functors \overline{G}: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}_{F/}} \} .$

Corollary 4.2.2.17.

• For any category $\operatorname{\mathcal{D}}$, the join functor

$\operatorname{Cat}\rightarrow \operatorname{Cat}_{\operatorname{\mathcal{D}}/ } \quad \quad \operatorname{\mathcal{C}}\mapsto \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$

admits a right adjoint, given on objects by the slice construction $(G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}) \mapsto \operatorname{\mathcal{E}}_{/G}$.

• For any category $\operatorname{\mathcal{C}}$, the join functor

$\operatorname{Cat}\rightarrow \operatorname{Cat}_{\operatorname{\mathcal{C}}/ } \quad \quad \operatorname{\mathcal{D}}\mapsto \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}$

admits a right adjoint, given on objects by the coslice construction $(F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}) \mapsto \operatorname{\mathcal{E}}_{F/}$.

Remark 4.2.2.18. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be a functor between categories. According to Remark 4.2.1.10, the slice category $\operatorname{\mathcal{E}}_{/G}$ can be identified with the iterated fiber product

$(\operatorname{Fun}( [0], \operatorname{\mathcal{E}}) \times _{ \operatorname{Fun}( \{ 0\} \times \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) } \operatorname{Fun}( [1] \times \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) ) \times _{ \operatorname{Fun}( \{ 1\} \times \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) } \{ G \} .$

Using Example 4.2.2.15, we can identify the left factor with the functor category $\operatorname{Fun}( \operatorname{\mathcal{D}}^{\triangleleft }, \operatorname{\mathcal{E}})$. We therefore obtain a pullback diagram of categories

$\xymatrix { \operatorname{\mathcal{E}}_{/G} \ar [r] \ar [d] & \operatorname{Fun}( \operatorname{\mathcal{D}}^{\triangleleft }, \operatorname{\mathcal{E}}) \ar [d] \\ \{ G\} \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}), }$

which recovers Example 4.2.2.11 at the level of objects.

Similarly, if $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}$ is a functor of categories, then the coslice category $\operatorname{\mathcal{E}}_{F/}$ fits into a pullback square

$\xymatrix { \operatorname{\mathcal{E}}_{F/} \ar [r] \ar [d] & \operatorname{Fun}( \operatorname{\mathcal{C}}^{\triangleright }, \operatorname{\mathcal{E}}) \ar [d] \\ \{ F\} \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}}). }$