Weak enriched Yoneda lemma

Let $\alpha :\mathcal{A}(K,-)\Rightarrow F$ be $\mathcal{V}$-natural. By [L4], its component at $K$ corresponds to an ordinary morphism ${\bar{\alpha }}_K:\mathcal{A}(K,K)\to FK$. Define $x_{\alpha }:={\bar{\alpha }}_K\circ j_K:\mathbf{1}\to FK$, where $j_K$ is the enriched identity of $K$. This is the evaluation map from the statement.

Conversely, let $x:\mathbf{1}\to FK$. For each object $A$, the functor structure of $F$ gives a morphism $F_{K,A}:\mathcal{A}(K,A)\to [FK,FA]$, and [L3] turns it into $F_{K,A}^{\mathrm{♯}}:\mathcal{A}(K,A)\otimes FK\to FA$. Compose with $1\otimes x$ and the right unitor to obtain ${\bar{\alpha }}_A^x:\mathcal{A}(K,A)\to FA$. By [L4], this determines a unique component ${\alpha }_A^x:\mathbf{1}\to [\mathcal{A}(K,A),FA]$.

The square criterion of [L2] for ${\alpha }^x$ is exactly the compatibility of $F_{K,-}$ with enriched composition: both sides become the same composite $\mathcal{A}(A,B)\otimes \mathcal{A}(K,A)\otimes FK\to FB$ after evaluating the internal homs and inserting $x$. So ${\alpha }^x$ is $\mathcal{V}$-natural.

Starting from $\alpha$, step 1.2 applied to $x_{\alpha }$ reconstructs the same family of components because the naturality square at $(K,A)$ sends the identity element $j_K$ to the value of ${\alpha }_A$ on $x_{\alpha }$. Starting from $x$, step 1.1 evaluates ${\alpha }_K^x$ at $j_K$ and recovers $x$ by definition. Hence the two constructions are inverse bijections.

Therefore evaluation at the enriched identity yields the stated bijection.