Defining a grading on tensor product of algebras Let $A$ be a commutative ring and $(\Delta_i)_{i\in I}$ a finite family of commutative monoids. Write $\Delta:=\prod_{i\in I}\Delta_i$. For each $i\in I$, let $E_i$ be a graded $A$-algebra of type $\Delta_i$ and $\varphi_i:\Delta_i\rightarrow\Delta$ the canonical homomorphism.  Then the author asserts the following:

Consider the $A$-module $E:=\bigotimes_{i\in I}E_i$; it is identified
with the direct sum of the submodules $E^\alpha$, where, for each
$\alpha\in\Delta$, we write $E^\alpha:=\bigotimes_{i\in
 I}E_i^{\alpha_i}$; the $E^\alpha$ therefore form a graduation of type
$\Delta$ on the $A$-module $E$.

I am not sure how we can derive the above-mentioned identification. We have
$$\bigotimes_{i\in I}E_i\cong\bigotimes_{i\in I}\bigoplus_{\lambda\in\Delta_i}E_i^{\lambda}.$$
But since $i$ appears in the indexing set of the direct sum, it is not possible to permutate the tensor and the direct sum. Any ideas on how we can derive
$$\bigotimes_{i\in I}E_i\cong\bigoplus_{\alpha\in\Delta}\bigotimes_{i\in I}E^{\alpha_i}_i?$$
 A: Here's an analogy for the calculation which might be helpful to think about. Imagine that we have the following finite collection of infinite formal sums
\begin{align*}
&x_1^0 + x_1^1 + x_1^2 + \cdots + x_1^i + \cdots\\
&x_2^0 + x_2^1 +  x_2^2 + \cdots + x_2^i + \cdots\\
&\qquad\qquad\qquad\vdots\\
&x_n^0 + x_n^1 + x_n^2 + \cdots +x_n^i + \cdots\\
\end{align*}
which we'd like to multiply together. That is, we'd like to find another way to express the product
$$
\prod_{i = 1}^n\sum_{j = 0}^\infty x_i^j = ( x_1^0 + x_1^1 + x_1^2 + \cdots )( x_2^0 +  x_2^1 + x_2^2 + \cdots)\cdots (x_n^0 + x_n^1 + x_n^2 + \cdots).
$$
How can we do this? We can write this as a sum of monomials! In such a sum, every term will be a product which itself consists of one term from each individual factor. Explicitly, for each element $\alpha = (\alpha_i)\in\Bbb{N}^n,$ we'll have a term $x^\alpha = x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}.$
Writing this all out, we have
$$
\prod_{i=1}^n\sum_{j = 0}^\infty x_i^j = \sum_{\alpha\in\Bbb{N}^n}\prod_{i = 1}^n x_i^{\alpha_i}.
$$
Notice how the formula here is nearly identical to the one you want to prove, except we've replaced $E_i^{\alpha_i}$ by $x_i^{\alpha_i}$ and $\Delta$ by $\Bbb{N}^n.$

Now let's examine the result you actually want to prove. The calculation will be nearly identical to the analogy I presented above, because $\otimes$ distributes over $\oplus.$
Suppose first for simplicity that we only have two monoids, $\Delta = \Delta_1$ and $\Delta' = \Delta_2,$ and that both are finite: $\Delta = \{e_1, e_2\},$ $\Delta' = \{e'_1, e'_2, e'_3\}.$ Then
$$
E = (E_1^{e_1}\oplus E_1^{e_2})\otimes(E_2^{e'_1}\oplus E_2^{e'_2}\oplus E_2^{e_3}).
$$
Since tensor products commute with limits (in particular, direct sums), we have
\begin{align*}
E &= (E_1^{e_1}\oplus E_1^{e_2})\otimes(E_2^{e'_1}\oplus E_2^{e'_2}\oplus E_2^{e_3})\\
&\cong \left(E_1^{e_1}\otimes(E_2^{e'_1}\oplus E_2^{e'_2}\oplus E_2^{e'_3})\right)\oplus\left(E_1^{e_2}\otimes(E_2^{e'_1}\oplus E_2^{e'_2}\oplus E_2^{e'_3})\right)\\
&\cong \left((E_1^{e_1}\otimes E_2^{e'_1})\oplus(E_1^{e_1}\otimes E_2^{e'_2})\oplus (E_1^{e_1}\otimes E_2^{e'_3})\right)\oplus\left((E_1^{e_2}\otimes E_2^{e'_1})\oplus(E_1^{e_2}\otimes E_2^{e'_2})\oplus (E_1^{e_2}\otimes E_2^{e'_3})\right)\\
&\cong (E_1^{e_1}\otimes E_2^{e'_1})\oplus(E_1^{e_1}\otimes E_2^{e'_2})\oplus (E_1^{e_1}\otimes E_2^{e'_3})\oplus(E_1^{e_2}\otimes E_2^{e'_1})\oplus(E_1^{e_2}\otimes E_2^{e'_2})\oplus (E_1^{e_2}\otimes E_2^{e'_3}).
\end{align*}
Notice that each summand here is a tensor product of one graded piece of $E_1$ and one graded piece of $E_2,$ and that each possible tensor of these graded pieces appears exactly once. This is what will happen in the general case, except we have a lot more to keep track of!
Let $\Delta_1,\dots,\Delta_n$ be our monoids. Let's also write $\Delta_i := \{e_j\mid j\in J_i\},$ so that $E_i = \bigoplus_{j\in J_i} E_i^{e_j}.$ Then we have
$$
E = \left(\bigoplus_{j\in J_1} E_1^{e_j}\right)\otimes\left(\bigoplus_{j\in J_2} E_2^{e_j}\right)\otimes\cdots\otimes\left(\bigoplus_{j\in J_n} E_n^{e_j}\right).
$$
Using the same reasoning as before, we may distribute the tensor products over the direct sums. For example, if we let $B = \left(\bigoplus_{j\in J_1} E_1^{e_j}\right)\otimes\left(\bigoplus_{j\in J_2} E_2^{e_j}\right)\otimes\cdots\otimes\left(\bigoplus_{j\in J_{n-1}} E_{n-1}^{e_j}\right),$ then we have
$$
E = B\otimes\left(\bigoplus_{j\in J_n} E_n^{e_j}\right) = \bigoplus_{j\in J_n} B\otimes E_n^{e_j}.
$$
Now, we continually apply this approach to distribute the tensor products over all the internal direct sums.
After $n$ steps, we'll end up with
$$
E = \bigoplus_{j_1\in J_1}\bigoplus_{j_2\in J_2}\cdots\bigoplus_{j_n\in J_n}\left(E_1^{e_{j_1}}\otimes E_2^{e_{j_2}}\otimes\cdots\otimes E_n^{e_{j_n}}\right).
$$
Letting $J = J_1\times J_2\times\cdots\times J_n,$ we can rewrite this as
$$
E = \bigoplus_{(j_1,j_2,\dots, j_n)\in J}\left(E_1^{e_{j_1}}\otimes E_2^{e_{j_2}}\otimes\cdots\otimes E_n^{e_{j_n}}\right).
$$
If $\alpha = (e_{j_1},e_{j_2},\dots, e_{j_n}),$ then $\left(E_1^{e_{j_1}}\otimes E_2^{e_{j_2}}\otimes\cdots\otimes E_n^{e_{j_n}}\right)$ is precisely $E^\alpha$ as defined in your original question, and $J$ is in bijection with $\Delta,$ with $j = (j_1,j_2,\dots, j_n)$ corresponding to $(e_{j_1},e_{j_2},\dots, e_{j_n}).$ So
$$
E = \bigoplus_{(j_1,j_2,\dots, j_n)\in J}\left(E_1^{e_{j_1}}\otimes E_2^{e_{j_2}}\otimes\cdots\otimes E_n^{e_{j_n}}\right) = \bigoplus_{\alpha\in\Delta} E^\alpha,
$$
as desired.
The key here is just repeatedly applying distributivity of $\otimes$ over $\oplus,$ and making sure to keep track of all your indices as you manipulate.
