If $C$ is a monoidal category, there is the canonical map $$ \operatorname{Hom}(A_1,A_2)\times\operatorname{Hom}(B_1,B_2)\to\operatorname{Hom}(A_1\otimes B_1,A_2\otimes B_2) $$ with $(f,g)\mapsto f\otimes g$.

If $\mathcal{C}$ is a category, then the category of functors $\mathcal{C}\to\mathcal{C}$ is a strict monoidal category where the bifunctor $\otimes$ is just composition of functors, and the units are both the identity functor.

In this case, what is the map on Hom-spaces? If $\eta\colon F_1\to F_2$ and $\zeta\colon G_1\to G_2$ are natural transformations of functors, what is the natural transformation $F_1\circ G_1\to F_2\circ G_2$?

For an object $X$ in $\mathcal{C}$, $\zeta_X\colon G_1(X)\to G_2(X)$ and $\eta_{G_1(X)}\colon F_1(G_1(X))\to F_2(G_1(X))$ but nothing seems to lead to a component $F_1(G_1(X))\to F_2(G_2(X))$.


1 Answer 1


$\require{AMScd}$ Pick natural transformations $\alpha:F_1\to F_2$ and $\beta:G_1\to G_2$. For any $X$ consider

$$\begin{CD} F_1G_1X @>\alpha_{G_1X}>> F_2G_1X \\@VV{F_1\beta_X}V @VV{F_2\beta_X}V\\ F_1G_2X @>{\alpha_{G_2X}}>> F_2G_2X \end{CD} $$

Naturality of $\alpha$ applied to the vertical maps $\beta_X:G_1X\to G_2X$ implies this is commutative.

Call the composition $(\alpha\circ\beta)_X:(F_1\circ G_1)X\to (F_2\circ G_2)X$.

  • 1
    $\begingroup$ Just to add for others: in a general monoidal category, $f \otimes g$ is by functoriality the same as $f \otimes B_2 \circ A_1 \otimes g = A_2 \otimes g \circ f \otimes B_1$. Applying this to natural transformations, the product is $\eta \circ \zeta = F_2 \zeta \circ \eta G_1 = \eta G_2 \circ F_1 \zeta$. The notation $\eta F$ and $F \eta$ are very standard and mean $(\eta F)_X = \eta_{F X}$ and $(F \eta)_X = F \eta_X$. (see whiskering on nlab. $\endgroup$
    – Jo Mo
    Feb 1, 2023 at 11:48

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