Associativity of Day convolution I'm trying to follow Day's argument to prove that $[\mathbf C,\mathbf{Sets}]$, where $\bf C$ is symmetric monoidal, is itself symmetric monoidal, but I'm stuck at the very beginning. Is there a way to prove that the convolution of two functors defines an associative "tensor" on $Psh(\mathbf C)$?
I apologize if the question seems too boring, please notice that


*

*Day's original paper is out of reach where I am;

*Google seems to give me nothing useful, 

*As you can see, the nlab page lacks the proof of associativity.


If I try to write down $F\star G\star H$ in the two ways, I feel I have to exploit some kind of Yoneda lemma written in coend form, but I'm not such an expert on ''endy'' notation to find such a form out.
Thanks!
 A: You are right that the Yoneda lemma plays the key role in the proof.
By the definition of convolution:
$$(F \otimes G) \otimes H \approx \int^{C, D} \left(\int^{A, B} F(A) \times G(B) \times \hom(C, A \otimes B)\right) \times H(D) \times \hom(-, C \otimes D)$$
Because products in $\mathbf{Set}$ preserve coends, the above is isomorphic to:
$$\int^{A, B, C, D} F(A) \times G(B) \times \hom(C, A \otimes B) \times H(D) \times \hom(-, C \otimes D)$$
The Yoneda lemma says that any covariant functor $K(A)$ is naturally isomorphic to $\hom(\hom(A, -), K)$. By definition, the last object is the end $\int_C K(C)^{\hom(A, C)}$. Therefore the Yoneda lemma says:
$$K(A) \approx \int_C K(C)^{\hom(A, C)}$$
By duality, we get the Yoneda lemma for contravariant functors $K$:
$$K(A) \approx \int^C K(C) \times \hom(A, C)$$
and from the perspective of the opposite category, for covariant functors $K$:
$$K(A) \approx \int^C K(C) \times \hom(C, A)$$
where the end turned into the coend and the exponent turned into the product.
In our case, we may reduce by Yoneda $\hom(C, A\otimes B)$ with $\hom(-, C \otimes D)$ under the coend:
$$\int^C \hom(-, C \otimes D) \times \hom(C, A\otimes B) \approx \hom(-, (A \otimes B) \otimes D)$$
obtaining:
$$\int^{A, B, D} F(A) \times G(B) \times H(D) \times \hom(-, (A \otimes B) \otimes D)$$
Due to associativity of $\otimes$ we know that $(A \otimes B) \otimes D \approx A \otimes (B \otimes D)$. Now it suffices to "unwind" the coend in the other direction.
A: The category of cocontinuous functors on $\widehat{C}$ with values in a cocomplete category is equivalent to the category of functors on $C$. In order to construct a natural isomorphism $(F \otimes G) \otimes H \cong F \otimes (G \otimes H)$ for $F,G,H \in \widehat{C}$, it is therefore enough to treat the case $F \in C$. Similarily, one reduces to $G \in C$ and $H \in C$. But then we use the associativity in $C$.
