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?

connection between composition and "inner composition"


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).

  • $\begingroup$ Arnaud, as far as I understand, $A$, $B$ and $\frac{C}{B}$ in this chain mean $1_A$, $1_B$ and $1_{\frac{C}{B}}$. $\endgroup$ – Sergei Akbarov Jun 13 at 18:41

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