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Let ${\mathcal C}$ be a symmetric closed monoidal category, and let me denote the internal hom-functor by a fraction $$ (X,Y)\mapsto\frac{Y}{X}, $$ so that we have an isomorphism of functors $$ \operatorname{Mor}(A\otimes B,C)\cong\operatorname{Mor}\left(A,\frac{C}{B}\right). $$ 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 corresponding properties.

I wonder if the following identity always holds $$ \bullet_{A,C,D}\circ\left(1_{\frac{D}{C}}\otimes\frac{\varphi}{1_A}\right)= \bullet_{A,B,D}\circ\left(\frac{1_D}{\varphi}\otimes1_{\frac{B}{A}}\right) $$ (for arbitrary objects $A,B,C,D$ and for arbitrary morphism $\varphi:B\to C$).

multiplication and division by a morphism under the inner multiplication

This is strange, I can prove this only in the case when the unit $I$ is a separating object in ${\mathcal C}$ (what does not always hold). Is it possible that there is a couterexample?

Edit: I asked this now at mathoverflow.

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    $\begingroup$ Where are you getting this "division" notation? $\endgroup$ – Derek Elkins Jun 23 at 0:59
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    $\begingroup$ @DerekElkins this is my own attempt to shorten the writings. Fraction seems to be a convenient notation for closed monoidal categories: en.wikipedia.org/wiki/… $\endgroup$ – Sergei Akbarov Jun 23 at 4:52
  • $\begingroup$ @ArnaudD. what do you think? $\endgroup$ – Sergei Akbarov Jun 23 at 9:51
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    $\begingroup$ The division notation is actually rather popular among allegory theorists and you can even find it in category theory introductions such as Fokkinga's works. $\endgroup$ – Musa Al-hassy Jun 23 at 12:37
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    $\begingroup$ Here's one of my favourites maartenfokkinga.github.io/utwente/mmf92b.pdf ;; it uses division for colimits ;; see "Categories, Allegories" for the use of division for residuals, which generalise Kan extensions ;-) $\endgroup$ – Musa Al-hassy Jun 26 at 13:14

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