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Stream: learning: questions

Topic: Enriching -> taking the underlying category = doing nothing

Marco Vianello (Jun 18 2026 at 07:28):

Let $\mathcal V$ be a closed monoidal caetgory, and write $[x, y]$ for the internal hom of $x$ and $y$. I'm trying to show that the underlying category of the enriched $\mathcal V$-category $\underline{\mathcal V}$ whose hom-objects are $\underline{\mathcal V}(x, y) = [x, y]$ is isomorphic to $\mathcal V$ itself. The composition in $\underline{\mathcal V}$ is defined as the transpose of the canonical map $[y, z]\otimes [x, y]\otimes x\to z$. The reference I'm following is Categorical Homotopy Theory by E .Riehl, Section 3.4.

I don't get where the proof proposed in the book is going. Let's agree to denote with $\underline{\mathcal V}_0$ the underlying ordinary category of $\underline{\mathcal V}$, so that

$$\underline{\mathcal V}_0(x, y) = \mathcal V(1, [x, y]) \cong \mathcal V(1\otimes x, y) \cong \mathcal V(x, y) \tag{1}$$

and the composition of $f\colon 1\to [x, y]$ with $g\colon 1\to [y, z]$ is given by

$$1\xrightarrow{\cong} 1\otimes 1\xrightarrow {g\otimes f} [y, z]\otimes [x, y]\xrightarrow{\underline\circ} [x, z] \tag{2}$$

where $\underline\circ$ is the composition operation of $\underline{\mathcal V}$.

Now, whenever $f\colon x\to y$ and $g\colon y \to z$ are ordinary arrows in $\mathcal V$ we consider the maps $\tilde f\colon 1\to [x ,y]$ and $\tilde g\colon 1\to [y, z]$ under the isomorphism $(1)$ above, and try to compose them. Then $(2)$ is adjunct to

$$1\otimes x\xrightarrow{\cong} 1\otimes 1\otimes s\xrightarrow {\tilde g\otimes \tilde f} [y, z]\otimes [x, y]\otimes x\xrightarrow{\underline\circ} [x, z]\otimes x\xrightarrow{\epsilon_z^x} z \tag{3}$$

since the adjunct of a map $a\to [x, b]$ is $a\otimes x\to [x, b]\otimes x\to b$, where the last arrow is the counit of the adjunction ${-\otimes x}\dashv {[x, -]}$. In particular, we can rewrite $(3)$ as

$$1\otimes x\xrightarrow{\cong} 1\otimes 1\otimes s\xrightarrow {\tilde g\otimes \tilde f} [y, z]\otimes [x, y]\otimes x\xrightarrow{\mathrm{id}\otimes \epsilon_y^x} [y, z]\otimes y\xrightarrow{\epsilon_z^y} z$$

(just look at how we defined $\underline\circ$). Modulo notation, this is the fourth formula at page 33 of prof. Riehl's book.

At this point I'm stuck. In particular, I suspect that there might be a few typos in the proof as printed in the book.

Kevin Carlson (Jun 18 2026 at 17:04):

I think this is the diagram with the key idea, which shows how you can get $g\circ f$ in the underlying category of $V$ as $\varepsilon_z\circ (\bar g\otimes f)\circ \lambda_x^{-1},$ where $\lambda$ is the unitor and $\eta,\varepsilon$ the unit and counit for $(-)\otimes y.$ We use that the transpose $\bar g$ is given by $1\stackrel{\eta_y,1}{\to}[y,1\otimes y]\stackrel{[y,g\circ\lambda_y]}{\to}[y,z]$ and bifunctoriality of $\otimes$ to set up the upper path in the diagram, then, chasing from upper-right to bottom-left, the naturality of $\varepsilon,$ the triangle identity, and naturality of $\lambda.$ Then the result you want comes repeating the same kind of argument in the other variable.

Marco Vianello (Jun 25 2026 at 15:46):

Ok, I think it's clear now. The main idea was basically that $\hat f$ is the unique map that makes the equation

$$\epsilon_x^y \circ (\hat f\otimes x) = f \circ \lambda_x$$

hold:)