Iterated Coproduct in a Monoidal Category; finding the unit of a monoid. Suppose $B$ is a monoidal category and further that the functors $-\bigotimes a:B\rightarrow B$ and $a\bigotimes -:B\rightarrow B$ preserve coproducts. The we have $\theta :\coprod _{b} a\bigotimes b_{n}\cong a\bigotimes \coprod _{n}b_{n}$, naturally in $a$, and similarly for the other functor. 
If we apply these isomorphisms twice, we get
$\coprod _{m} a^{m}\bigotimes \coprod _{n} a^{n}\cong \coprod _{m}\left ( a^{m} \bigotimes \coprod _{n} a^{n}\right )\cong \coprod_{m} \left ( \coprod_{n} \left ( a^{m}\bigotimes a^{n} \right )  \right )$.
Actually the latter isomorphism $\coprod _{m} \theta $, isn't it ? After all, $\coprod $ is just a special type of limit functor. 
I want to find an explicit isomorphism between the latter coproduct and 
$\coprod _{n,m} a^{m}\bigotimes a^{n}$. 
I need this to prove that $\eta  _{a}$, the unit of the monoid $\coprod _{n} a^{n}$ is given by the injection $i_{0}:e\rightarrow \coprod _{n} a^{n}$. I have $\mu $ and am trying to show that $\mu \circ (\eta \bigotimes 1)=\lambda $. All this works out very easily in $Set$ but I am having trouble with the general case. 
 A: So we take $\eta:e\to\coprod_n a^n$ to be the inclusion of $e=a^0$ into the coproduct. We want to show that $\mu(\eta\otimes1)=\lambda:e\otimes\coprod_n a^n\to\coprod_n a^n$. Consider the following commutative diagram
$$
\begin{matrix}
 & & e⊗∐_n a^n & → & ∐_m a^m⊗∐_n a^n & → & ∐_{m,n}a^m⊗a^n & → & ∐_k a^k \\
 & & \uparrow & & \uparrow & & \uparrow \\
e⊗a^n &\hookrightarrow & ∐_n e⊗a^n &→&∐_n(∐_m a^m)⊗a^n &→&∐_n∐_m(a^m⊗a^n)
\end{matrix}
$$
The arrow we are interested in is the upper path $\mu(\eta⊗1)$. I recommend you compose it with the arrow $1_e⊗i_n:e⊗a^n\to e⊗\coprod_n a^n$ and try to show that this is just $\lambda(1_e⊗i_n)$. Come back if you encounter problems.

Here is the solution: Composing the upper row with $1_e⊗i_n:e⊗a^n→e⊗∐_na^n$ gives the lower row augmented by the map $∐_n∐_m(a^m⊗a^n)→∐_k a^k$. The first two maps of the lower row compose to the map $\eta⊗1_{a^n}$. Composed with the third map, this gives $i_n\circ i_0:a^0⊗a^n \to ∐_m(a^m⊗a^n) \to ∐_n∐_m(a^m⊗a^n)$, which then composed with the upward arrow yields $i_{0,n}$. Composing with the last arrow produces the map $i_n\circ\lambda$, which by naturality of $\lambda$ is the same as $\lambda(1_e⊗i_n)$.
