Why is a closed monidal category enriched over itself?

Question

Let $M$ be a left-closed monidal category (Assume more such as symmetry if needed). Let $i$ be the unity object of $M$. Let $\alpha,\lambda,\rho$ be the associator, left unitor, right unitor of $M$, respectively. Let $\Phi$ be the currying of $M$ (so that $\Phi_{a,b,c}:(b\otimes a,c)\to(a,[b,c])$), where $(x,y)$ is the hom set $\hom(x,y)$. For each pair of objects $a,b$ in $M$, let $\epsilon_{a,b}:a\otimes[a,b]\to b$ be the the morphism $\Phi\left(1_{[a,b]}\right)$. For each object $a$, define $\boldsymbol{1}_a:i\to[a,a]$ by \begin{equation} \boldsymbol{1}_a = \Phi\left(a\otimes i\xrightarrow{\rho_a}a\right)\text{.} \end{equation} For each triple of objects $a,b,c$, define $\square_{a,b,c}:[a,b]\otimes[b,c]\to[a,c]$ by \begin{equation} \square_{a,b,c} = \Phi\left(a\otimes\left([a,b]\otimes[b,c]\right)\xrightarrow{\alpha_{a,[a,b],[b,c]}}\left(a\otimes[a,b]\right)\otimes[b,c]\xrightarrow{\epsilon_{a,b}\otimes 1_{[b,c]}}b\otimes[b,c]\xrightarrow{\epsilon_{b,c}}c\right)\text{.} \end{equation}

Now I want to show that, with $\boldsymbol{1}$ and $\square$ defined above, $M$ is enriched over $M$. What I need to show is that the following three diagrams (called coherence conditions for enriched categories) commute:

In section 1.6 of Kelly's "Basic Concepts of Enriched Category Theory," (http://www.tac.mta.ca/tac/reprints/articles/10/tr10.pdf), it is shown, using the naturality of $\Phi$, that \begin{equation} \epsilon_{a,c}\circ\left(1_a\otimes \square_{a,b,c}\right) = \epsilon_{b,c} \circ \left(\epsilon_{a,b}\otimes 1_{[b,c]}\right) \circ \alpha_{a,[a,b],[b,c]}\text{.} \end{equation} And it says that now it is easy to complete the proof. But I do not know how to proceed.

Answer 1

I figured it out. I consulted the comments, and the following post: A closed symmetric monoidal category is enriched over itself.

This answer might be uglier than it could have been. Any new answer is welcome.

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Important Note. I write $(x)f$ instead of $f(x)$, and write $f\circ g$ instead of $g\circ f$. This does not apply to the question, but just to this answer.

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Proof. Using the fact that $\Phi$ is natural over all three arguments, one can somehow show that, for all morphisms $g:b'\to b,\ h:c\to c',\ \varphi:b\otimes a\to c$ in $M$, it holds that \begin{equation} \left(g\otimes\left(\varphi\Phi\right)\right)\circ\epsilon_{b,c}=\left(g\otimes 1_a\right)\circ\varphi\text{.}\tag{0} \end{equation}

We will use the above equation to prove the coherence conditions for enriched categories.

(First Coherence Condition). Let $\alpha=\alpha_{[a,b],[b,c],[c,d]}$. By the naturality of $\Phi$, the following diagram commutes: The following computation implies the first coherence condition. \begin{align*} &\left(\alpha\circ\left(\square_{a,b,c}\otimes 1_{[c,d]}\right)\circ\square_{a,c,d}\right)\Phi^{-1}\\ =& \left(1_a\otimes\alpha\right)\circ\left(1_a\otimes\left(\square_{a,b,c}\otimes 1_{[c,d]}\right)\right)\circ\alpha_{a,[a,c],[c,d]}\\ &\circ\left(\epsilon\otimes 1_{[c,d]}\right)\circ\epsilon_{c,d} && \text{(Diagram (1) top two squares)}\\ =& \left(1_a\otimes\alpha\right)\circ\alpha_{a,[a,b]\otimes[b,c],[c,d]}\circ\left(\left(1_a\otimes\square_{a,b,c}\right)\otimes 1_{[c,d]}\right)\\ & \circ\left(\epsilon_{a,c}\otimes1_{[c,d]}\right)\circ\epsilon_{c,d} && \text{($\alpha$ naturality)}\\ =& \left(1_a\otimes\alpha\right)\circ\alpha_{a,[a,b]\otimes[b,c],[c,d]}\circ\left(\left(\left(1_a\otimes\square_{a,b,c}\right)\circ\epsilon_{a,c}\right)\otimes1_{[c,d]}\right)\\ & \circ\epsilon_{c,d}\\ =& \left(1_a\otimes\alpha\right)\circ\alpha_{a,[a,b]\otimes[b,c],[c,d]}\circ\left(\alpha_{a,[a,b],[b,c]}\otimes1_{[c,d]}\right)\\ & \circ\left(\left(\epsilon_{a,b}\otimes1_{[b,c]}\right)\otimes1_{[c,d]}\right)\circ\left(\epsilon_{b,c}\otimes1_{[c,d]}\right)\circ\epsilon_{c,d} && \text{(Equation (0))}\\ =& \alpha_{a,[a,b],[b,c]\otimes[c,d]}\circ\alpha_{a\otimes[a,b],[b,c],[c,d]}\circ\left(\left(\epsilon_{a,b}\otimes1_{[b,c]}\right)\otimes1_{[c,d]}\right)\\ & \circ\left(\epsilon_{b,c}\otimes1_{[c,d]}\right)\circ\epsilon_{c,d} && \text{($\alpha$ pentagon)}\\ =& \alpha_{a,[a,b],[b,c]\otimes[c,d]}\circ\left(\epsilon_{a,b}\otimes1_{[b,c]\otimes[c,d]}\right)\circ\alpha_{b,[b,c],[c,d]}\\ & \circ\left(\epsilon_{b,c}\otimes1_{[c,d]}\right)\circ\epsilon_{c,d} && \text{($\alpha$ naturality)}\\ =& \alpha_{a,[a,b],[b,c]\otimes[c,d]}\circ\left(\epsilon_{a,b}\otimes1_{[b,c]\otimes[c,d]}\right)\circ\left(1_b\otimes\square_{b,c,d}\right)\circ\epsilon_{b,d} && \text{(Equation (0))}\\ =& \alpha_{a,[a,b],[b,c]\otimes[c,d]}\circ\left(\epsilon_{a,b}\otimes\square_{b,c,d}\right)\circ\epsilon_{b,d}\\ =& \alpha_{a,[a,b],[b,c]\otimes[c,d]}\circ\left(1_{a\otimes[a,b]}\otimes\square_{b,c,d}\right)\circ\left(\epsilon_{a,b}\otimes1_{[b,d]}\right)\circ\epsilon_{b,d}\\ =& \left(1_a\otimes\left(1_{[a,b]}\otimes\square_{b,c,d}\right)\right)\circ\alpha_{a,[a,b],[b,d]}\circ\left(\epsilon_{a,b}\otimes1_{[b,d]}\right)\circ\epsilon_{b,d} && \text{($\alpha$ naturality)}\\ =& \left(\left(1_{[a,b]}\otimes\square_{b,c,d}\right)\circ\square_{a,b,d}\right)\Phi^{-1}\text{.} && \text{(Diagram (1) bottom square)} \end{align*}

(Second Coherence Condition). By the naturality of $\Phi$, the following diagram commutes: The following computation implies the second coherence condition. \begin{align*} & \left(\left(\boldsymbol{1}_a\otimes1_{[a,b]}\right)\circ\square_{a,a,b}\right)\Phi^{-1}\\ =& \left(1_a\otimes\left(\boldsymbol{1}_a\otimes1_{[a,b]}\right)\right)\circ\alpha_{a,[a,a],[a,b]}\circ\left(\epsilon_{a,a}\otimes1_{[a,b]}\right)\circ\epsilon_{a,b} && \text{(Diagram (2) top square)}\\ =& \alpha_{a,i,[a,b]}\circ\left(\left(1_a\otimes\boldsymbol{1}_a\right)\otimes1_{[a,b]}\right)\circ\left(\epsilon_{a,a}\otimes1_{[a,b]}\right)\circ\epsilon_{a,b} && \text{($\alpha$ naturality)}\\ =& \alpha_{a,i,[a,b]}\circ\left(\left(\left(1_a\otimes\boldsymbol{1}_a\right)\circ\epsilon_{a,a}\right)\otimes1_{[a,b]}\right)\circ\epsilon_{a,b}\\ =& \alpha_{a,i,[a,b]}\circ\left(\rho_a\otimes1_{[a,b]}\right)\circ\epsilon_{a,b} && \text{(Equation (0))}\\ =& \left(1_a\otimes\lambda_{[a,b]}\right)\circ\epsilon_{a,b} && \text{($\lambda,\rho$ triangle)}\\ =& \lambda_{[a,b]}\Phi^{-1}\text{.} && \text{(Diagram (2) bottom square)} \end{align*}

(Third Coherence Condition). By the naturality of $\Phi$, the following diagram commutes: The following computation implies the third coherence condition. \begin{align*} & \left(\left(1_{[a,b]}\otimes\boldsymbol{1}_b\right)\circ\square_{a,b,b}\right)\Phi^{-1}\\ =& \left(1_a\otimes\left(1_{[a,b]}\otimes\boldsymbol{1}_b\right)\right)\circ\alpha_{a,[a,b],[b,b]}\circ\left(\epsilon_{a,b}\otimes1_{[b,b]}\right)\circ\epsilon_{b,b} && \text{(Diagram (3) top square)}\\ =& \alpha_{a,[a,b],i}\circ\left(1_{a\otimes[a,b]}\otimes\boldsymbol{1}_b\right)\circ\left(\epsilon_{a,b}\otimes1_{[b,b]}\right)\circ\epsilon_{b,b} && \text{($\alpha$ naturality)}\\ =& \alpha_{a,[a,b],i}\circ\left(\epsilon_{a,b}\otimes\boldsymbol{1}_b\right)\circ\epsilon_{b,b}\\ =& \alpha_{a,[a,b],i}\circ\left(\epsilon_{a,b}\otimes1_i\right)\circ\rho_b && \text{(Equation (0))}\\ =& \alpha_{a,[a,b],i}\circ\rho_{a\otimes[a,b]}\circ\epsilon_{a,b} && \text{($\rho$ naturality)}\\ =& \left(1_a\otimes\rho_{[a,b]}\right)\circ\epsilon_{a,b} && \text{($\rho$ triangle)}\\ =& \rho_{[a,b]}\Phi^{-1}\text{.} && \text{(Diagram (3) bottom square)} \end{align*} This completes the proof.

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Remarks.

1. Any left closed monoidal category is self-enriched. No assumptions like braidedness or symmetry is needed.
2. The above proof does not work for right closed monoidal categories. The coherence conditions does not even make sense. I think this will not be an issue if we assume a braiding of $M$. If one wants to work without a braiding, then what seems the most appropriate is that one should define left enrichedness and right enrichedness separately, and claim that left closed monoidal categories are self left enriched, and right closed ones are self right enriched.