Let $\mathcal{V}$ be a closed symmetric monoidal category. One is supposed to show that if $\underline{C}$ is an enriched category, $x, y \in \underline{C}$ are objects, then if they are isomorphic in the underlying category $C$ of $\underline{C}$, the representable $\mathcal{V}$-functors $$\underline{C}(x,-), \underline{C}(y,-): \underline{C} \rightarrow \underline{\mathcal{V}}$$ are $\mathcal{V}$-isomorphic. I have tried to construct a natural isomorphism of enriched functors, but i am having real problems with showing that the construction really is an enriched natural transformation. Any hints or solutions?

  • $\begingroup$ You need to start with the enriched composition of $\underline{\mathcal{C}}$ and restrict along the morphism $I \to \underline{\mathcal{C}}(x, y)$ that picks out the particular isomorphism $x \to y$ you are given. $\endgroup$ – Zhen Lin Mar 2 '14 at 17:10
  • $\begingroup$ @ZhenLin Right, that is what I started with - but how to show this is enriched natural? $\endgroup$ – user101036 Mar 2 '14 at 17:20
  • $\begingroup$ Use enriched associativity, of course. $\endgroup$ – Zhen Lin Mar 2 '14 at 18:25
  • $\begingroup$ Would you mind elaborating? I have filled 12 pages of commutative diagrams so it would be greatly appreciated. $\endgroup$ – user101036 Mar 2 '14 at 18:28
  • $\begingroup$ @ZhenLin Or just a sketch would be appreciated when and if you have the time! $\endgroup$ – user101036 Mar 2 '14 at 20:12

Commutative diagrams are hard to draw here, so let me outline the steps in words instead. Suppose $f : x \to y$ and $g : y \to x$ are mutually inverse morphisms in the underlying category of $\underline{\mathcal{C}}$. For ease of notation I will pretend $\mathcal{V}$ is a strict monoidal category.

  1. Recall that the morphisms $a \to b$ in the underlying category of $\underline{\mathcal{C}}$ are defined to be morphisms $I \to \underline{\mathcal{C}} (a, b)$ in $\mathcal{V}$, with the evident induced composition. Writing $\mu : \underline{\mathcal{C}} (b, c) \otimes \underline{\mathcal{C}} (a, b) \to \underline{\mathcal{C}} (a, c)$ for the composition in $\underline{\mathcal{C}}$ and $\eta_a : I \to \underline{\mathcal{C}} (a, a)$ for the identity morphisms, that means $\mu \circ (f \otimes g) = \eta_y$ and $\mu \circ (g \otimes f) = \eta_x$.
  2. Consider $(g^*)_z = \mu \circ (g \otimes \mathrm{id}) : \underline{\mathcal{C}} (x, z) \to \underline{\mathcal{C}} (y, z)$ and $(f^*)_z = \mu \circ (f \otimes \mathrm{id}) : \underline{\mathcal{C}} (y, z) \to \underline{\mathcal{C}} (x, z)$. By enriched associativity, $(g^*)_z \circ (f^*)_z = (\eta_y^*)_z$ and $(f^*)_z \circ (g^*)_z = (\eta_x^*)_z$, and enriched unitality means $(\eta_x^*)_z = \mathrm{id}$ and $(\eta_y^*)_z = \mathrm{id}$, so $(f^*)_z$ and $(g^*)_z$ are indeed isomorphisms in $\mathcal{V}$.
  3. Finally, we must show enriched naturality of $(f^*)_z$ and $(g^*)_z$ in $z$; by symmetry it suffices to consider just $(f^*)_z$. The claim is simply that, for all $w$, we have $\mu \circ (\mathrm{id} \otimes (f^*)_z) = (f^*)_w \circ \mu$ as morphisms $\underline{\mathcal{C}} (z, w) \otimes \underline{\mathcal{C}} (y, z) \to \underline{\mathcal{C}} (y, w)$. This again is a consequence of enriched associativity.
  • $\begingroup$ I am fine with 1 and 2, but I don't see why 3 coincides with the definition for the for an enriched natural transformation. On pg. 35 , def . 3.5.8 of math.harvard.edu/~eriehl/cathtpy.pdf one defines it as a certain commutative square. When we're in V, I see that we can use adjointness to simplify it somewhat, so I agree that in that diagram, first going right, then down (i.e $(\alpha_y)_\ast \circ F_{x,y})$ is the same by adjointness to $f^\ast_w \circ \mu$. But why is the other direction what you wrote? I am sure it's elementary, but I am a bit confused. $\endgroup$ – user101036 Mar 3 '14 at 12:58
  • $\begingroup$ And I just want to add that I am very grateful for the time you are taking and sharing your knowledge. From what I understand, $G_{x,y}$ in our case is the transpose of the composition morphism. It probably follows by adjointness that it transforms as you said but I can't see why. $\endgroup$ – user101036 Mar 3 '14 at 12:59
  • $\begingroup$ Yes, use the tensor–hom adjunction to reduce it to my equation. $\endgroup$ – Zhen Lin Mar 3 '14 at 13:54
  • $\begingroup$ Sorry, but why does it reduce? I realize this should follow easily but I can't see why as in the diagram $(f)^\ast_z \circ G_{x,y}$ where $G_{x,y}$ is adjoint to composition reduces under the adjunction to $\mu \otimes (id \otimes (f^\ast)_z)$. $\endgroup$ – user101036 Mar 3 '14 at 17:01
  • $\begingroup$ This is a fact about two-variable adjunctions (in ordinary categories). Try working out in the case $\mathcal{V} = \mathbf{Set}$ first. $\endgroup$ – Zhen Lin Mar 3 '14 at 17:31

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