Unit of Day convolution of functors Let $(\mathcal{C}, \otimes_\mathcal{C}, 1_\mathcal{C})$ be a small symmetric monoidal category, and let $(\mathcal{D}, \otimes_\mathcal{D}, 1_\mathcal{D})$ be a symmetric monoidal category. For simplicity let us assume that $\mathcal{D}$ is bicomplete. I am trying to prove that the functor category $\mathrm{Fun}(\mathcal{C}, \mathcal{D})$ admits a symmetric monoidal structure coming from Day convolution of functors.
Let $X, Y \in \mathrm{Fun}(\mathcal{C}, \mathcal{D})$ and let $\bar{\otimes}$ denote the external tensor product of $X$ and $Y$. The Day convolution $X \otimes_{\mathrm{Day}} Y$ of $X$ and $Y$ is then defined as the left Kan extension of the external tensor product of $X$ and $Y$ along $\otimes_\mathcal{C}$. Using the adjunction
$$
\mathrm{Fun}(\mathcal{C}, \mathcal{D})(X \otimes_{\mathrm{Day}} Y, Z) \cong \mathrm{Fun}(\mathcal{C} \times \mathcal{C}, \mathcal{D})(X \bar{\otimes} Y, Z \circ \otimes_\mathcal{C})
$$
and Yoneda I was able to prove that Day convolution is symmetric, and I think that associativity can be proven in the same way. However, I want to show that the functor $\mathcal{C}(1_\mathcal{C}, -) \colon \mathcal{C} \to \mathcal{D}$ is the unit of $\otimes_{\mathrm{Day}}$. Is there a proof of this which is similar to the strategy just described that does not use the coend description of Day convolution?
Thanks.  
EDIT: 
At the moment I am trying to give a proof using the characterization of Day convolution as a coend but I cannot quite make it. We have that
\begin{align*}
X \otimes_{\mathrm{Day}} \mathcal{C}(1_C,-) &\cong \int^{c_1, c_2} \mathcal{C}(c_1 \otimes_C c_2,-) \otimes_D X(c_1) \otimes_D \mathcal{C}(1_C, c_2) \\
&\cong \int^{c_1}\int^{c_2} \mathcal{C}(c_1 \otimes_C c_2,-) \otimes_D X(c_1) \otimes_D \mathcal{C}(1_C, c_2)
\end{align*}
At this moment I want to use the co-Yoneda lemma but I cannot figure out how to do this in a way that gets us where we want. Thanks.
 A: We have to assume that that the tensor product of $D$ commutes with colimits in each variable (i.e. that $(D,\otimes,1)$ is a cocomplete symmetric monoidal category). Of course we do not need that $D$ has limits.
If $c \in C$, then $C(c,-) : C \to \mathsf{Set}$ is not a functor to $\mathcal{D}$ as you write, but we may compose it with the unique cocontinuous symmetric monoidal functor $\mathsf{Set} \to D$, namely $X \mapsto X \otimes 1 = 1 ^{\oplus X}$.
If $X : C \to D$ is a functor and $c \in C$, then we compute
$$X \otimes_{Day} (C(c,-) \otimes 1) = \int^{a,b \in C} C(a \otimes b,-) \otimes X(a) \otimes (C(c,b) \otimes 1)$$
$$=\int^{a \in C} X(a) \otimes \int^{b \in C} C(a \otimes b,-) \otimes C(c,b) =\int^{a \in C} X(a) \otimes C(a \otimes c,-).$$
The last step was the Yoneda Lemma (in coend form). Next, if $c=1$, we can simplify this to $X$ by another application of the Yoneda Lemma (in coend form).
The same calculation can be made in the language of Kan extensions as requested:
$$Fun(C,D)(X \otimes_{Day} C(1,-) \otimes 1,Z)=Fun(C \times C,D)(X \overline{\otimes} C(1,-) \otimes 1,Z \circ \otimes)$$
$$ = \int_{a,b \in C} D(X(a) \otimes C(1,b),Z(a \otimes b)) = \int_{a,b \in C} [C(1,b),D(X(a),Z(a \otimes b))]$$
$$=\int_{a \in C} D(X(a),Z(a))=Fun(C,D)(X,Z)$$
The second to last step was the Yoneda Lemma (in end form).
