2
$\begingroup$

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"

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.