4.3.2 Joins of Categories
Our next goal is to characterize the slice categories of Construction 4.3.1.8 by a universal mapping property.
Definition 4.3.2.1 (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
\[ \operatorname{Hom}_{\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}}(X,Y) = \begin{cases} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) & \textnormal{if } X,Y \in \operatorname{Ob}(\operatorname{\mathcal{C}}) \\ \operatorname{Hom}_{\operatorname{\mathcal{D}}}(X,Y) & \textnormal{if } X,Y \in \operatorname{Ob}(\operatorname{\mathcal{D}}) \\ \ast & \textnormal{if } X \in \operatorname{Ob}(\operatorname{\mathcal{C}}), Y \in \operatorname{Ob}(\operatorname{\mathcal{D}}) \\ \emptyset & \textnormal{if } X \in \operatorname{Ob}(\operatorname{\mathcal{D}}), Y \in \operatorname{Ob}(\operatorname{\mathcal{C}}). \end{cases} \]
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 $\operatorname{\mathcal{D}}$.
Example 4.3.2.4. 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 4.3.2.5 (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 $\operatorname{\mathcal{C}}^{\triangleright }$).
Warning 4.3.2.8. The join operation of Definition 4.3.2.1 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 4.3.2.1 with the slice categories of Construction 4.3.1.8. We begin with a simple observation.
Lemma 4.3.2.9. 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 4.3.1.11, 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 4.3.2.10. 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 4.3.2.11. Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be a functor of categories. Applying Proposition 4.3.2.10 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 4.3.2.12. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{E}}$ be categories and let $S$ be an object of $\operatorname{\mathcal{E}}$. Applying Proposition 4.3.2.10 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 4.3.2.10, we can use Remark 4.3.1.11 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 4.3.2.10 in a more symmetric fashion:
Proposition 4.3.2.13. 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 4.3.2.9. Then evaluation on $u$ induces a bijection
\[ \xymatrix@R =50pt@C=50pt{ \{ \textnormal{Functors $U: \operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ with $U|_{\operatorname{\mathcal{C}}} = F$ and $U|_{\operatorname{\mathcal{D}}} = G$} \} \ar [d] \\ \{ \textnormal{Natural transformations from $F \circ \pi _{\operatorname{\mathcal{C}}}$ to $G \circ \pi _{\operatorname{\mathcal{D}}}$ } \} }. \]
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 $U(t \circ s) = G(t \circ s) = G(t) \circ G(s) = U(t) \circ U(s)$.
$\square$
Example 4.3.2.15 (The Universal Property of a Cone). Let $\operatorname{\mathcal{C}}$ be a category. Applying Remark 4.3.2.14 in the special case $\operatorname{\mathcal{D}}= [0]$, we obtain a pushout diagram of categories
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\times \{ 1\} \ar [r] \ar [d] & \operatorname{\mathcal{C}}\times {[1]} \ar [d] \\ {[0]} \ar [r] & \operatorname{\mathcal{C}}^{\triangleright }, } \]
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 4.3.2.11 and 4.3.2.12. 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
\[ \xymatrix@R =50pt@C=50pt{ \{ 0\} \times \operatorname{\mathcal{D}}\ar [r] \ar [d] & [1] \times \operatorname{\mathcal{D}}\ar [d] \\ {[0]} \ar [r] & \operatorname{\mathcal{D}}^{\triangleleft }. } \]
For completeness, we record the dual of Proposition 4.3.2.10, which supplies a universal property of coslice categories (and is also a reformulation of Proposition 4.3.2.13).
Corollary 4.3.2.16. 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 4.3.2.17.
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 $(F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{E}}) \mapsto \operatorname{\mathcal{E}}_{F/}$.