# Connection between composition and “inner composition” in closed monoidal categories

Let $${\mathcal C}$$ be a symmetric closed monoidal category, $$I$$ its unit object, $$\lambda_X:I\otimes X\to X$$ the left unit morphism, and let me denote the internal hom-functor by a fraction $$(X,Y)\mapsto\frac{Y}{X},$$ so that we have a natural isomorphism of functors $$\eta_{A,B,C}:\operatorname{Mor}(A\otimes B,C)\to \operatorname{Mor}\left(A,\frac{C}{B}\right).$$

This bijection, in particular, assigns to each morphism $$\varphi:X\to Y$$ a morphism $$\widehat{\varphi}=\eta_{I,X,Y}(\varphi\circ\lambda_X):I\to\frac{Y}{X}$$

Further, as is known, $${\mathcal C}$$ is an enriched category over itself. For each objects $$A,B,C$$ let me denote by $$\bullet_{A,B,C}$$ the "inner composition" in $${\mathcal C}$$ as in an enriched category, i.e. the morphism $$\bullet_{A,B,C}:\frac{C}{B}\otimes\frac{B}{A}\to\frac{C}{A}$$ with the necessary properties.

I think this "inner composition" $$\bullet_{A,B,C}$$ must be connected with the usual composition $$\circ$$ of morphisms by the identity $$\widehat{\psi\circ\varphi}=\bullet_{A,B,C}\circ (\widehat{\psi}\otimes\widehat{\varphi})\circ\lambda_I^{-1}$$ for each $$\varphi:A\to B$$ and $$\psi:B\to C$$. But I don't understand how people prove this.

I think there is a trick that I don't know. Can anybody enlighten me?

Let $$\epsilon^{A}_B$$ be the map $$\eta_{\frac{B}{A},A,B}^{-1}(1_{\frac{B}{A}})$$, or in other words, let $$\epsilon^{A}$$ be the counit of the adjunction $$? \otimes_A\dashv \frac{??}{A}$$, and similarly for $$B,C$$. Then the inverse of $$\eta_{A,B,C}$$ is the map $$g\mapsto \epsilon^{B}_C \circ (g\otimes B)$$. Moreover, the inner composition $$\frac{C}{B}\otimes\frac{B}{A}\to\frac{C}{A}$$ is the image of the composite $$\frac{C}{B}\otimes\frac{B}{A}\otimes A\stackrel{\frac{C}{B}\otimes \epsilon_{B}^{A}}{\longrightarrow} \frac{C}{B}\otimes B \stackrel{ \epsilon_{C}^B}{\longrightarrow} C$$ under the bijection $$\eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}$$.
As a consequence, and because of the naturality of $$\eta$$, we find that \begin{align}\bullet_{A,B,C}\circ (\widehat{\psi}\otimes\widehat{\varphi}) & = \eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \right) \circ (\widehat{\psi}\otimes \widehat{\varphi})\\ & = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \circ (\widehat{\psi}\otimes \widehat{\varphi}\otimes A)\right) \\ & = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\widehat{\psi}\otimes (\epsilon^{A}_B \circ ( \widehat{\varphi}\otimes A))\right) \right) \\ & = \eta_{I\otimes I,A,C}\left(\epsilon^B_C \circ \left(\widehat{\psi}\otimes (\varphi\circ \lambda_A) \right) \right) \\ & = \eta_{I,A,C}\left(\epsilon^B_C \circ (\widehat{\psi}\otimes B) \circ (I\otimes (\varphi\circ \lambda_A)) \right) \\ & = \eta_{I\otimes I,A,C}\left(\psi\circ \lambda_B \circ (I\otimes (\varphi\circ \lambda_A)) \right)\\ & = \eta_{I\otimes I,A,C}\left(\psi\circ \varphi\circ \lambda_A \circ \lambda_{I\otimes A} \right) \\ & = \eta_{I\otimes I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \circ (\lambda_{I}\otimes A) \right)\\ & = \eta_{I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \right)\circ \lambda_{I} \\ & = \widehat{\psi\circ \varphi} \circ \lambda_I\end{align} (with some associators missing, but it should work).
• Arnaud, as far as I understand, $A$, $B$ and $\frac{C}{B}$ in this chain mean $1_A$, $1_B$ and $1_{\frac{C}{B}}$. – Sergei Akbarov Jun 13 at 18:41