The horizontal composition1,2 of two natural transformations $\alpha \colon F\Longrightarrow G$ and $\beta \colon H\Longrightarrow K$ as in the diagram

of $\alpha $ and $\beta $ is the natural transformation

\[ \beta \mathbin {\star }\alpha \colon \webleft (H\circ F\webright )\Longrightarrow \webleft (K\circ G\webright ), \]

as in the diagram

consisting of the collection

\[ \webleft\{ \webleft (\beta \mathbin {\star }\alpha \webright )_{A} \colon H\webleft (F\webleft (A\webright )\webright ) \to K\webleft (G\webleft (A\webright )\webright ) \webright\} _{A\in \text{Obj}\webleft (\mathcal{C}\webright )}, \]

of morphisms of $\mathcal{E}$ with


1Further Terminology: Also called the Godement product of $\alpha $ and $\beta $.
2Horizontal composition forms a map

\[ \mathbin {\star }_{\webleft (F,H\webright ),\webleft (G,K\webright )}\colon \text{Nat}\webleft (H,K\webright )\times \text{Nat}\webleft (F,G\webright )\to \text{Nat}\webleft (H\circ F,K\circ G\webright ). \]

First, we claim that we indeed have

This is, however, simply the naturality square for $\beta $ applied to the morphism $\alpha _{A}\colon F\webleft (A\webright )\to G\webleft (A\webright )$. Next, we check the naturality condition for $\beta \mathbin {\star }\alpha $, which is the requirement that the boundary of the diagram

commutes. Since

  1. Subdiagram (1) commutes by the naturality of $\alpha $.
  2. Subdiagram (2) commutes by the naturality of $\beta $.

so does the boundary diagram. Hence $\beta \circ \alpha $ is a natural transformation.1


1Reference: Proposition 1.3.4 of [Borceux, Handbook of Categorical Algebra I].


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