# Kerodon

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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 $U(t \circ s) = G(t \circ s) = G(t) \circ G(s) = U(t) \circ U(s)$.

$\square$