I want to prove that $Mat_{n_1}(k) \otimes_k Mat_{n_2}(k) \cong Mat_{n_1n_2}(k) $ (as $k$-algebras) where $k$ is a field by checking the universal property.
Namely, I need to inclusion $Mat_{n_{1,2}}(k) \hookrightarrow Mat_{n_1n_2}(k)$ -- that must be obvious.I can put $n_1 \times n_1$ matrix to the up-left and $n_2 \times n_2$ to the down-right.
Now I need to show that for any $k$-algebra $A$ and for any $\phi_i: Mat_{n_i}(k) \to A $, such that $\phi_1(A)$ commures with $\phi_2(B)$ for any $A \in Mat_{n_{1}}(k), \; B \in Mat_{n_{2}}(k) $ there is a unique $\Phi: Mat_{n_1n_2}(k) \to A$ such that $\Phi \circ \mathcal{i}_i = \phi_i. $
In other words, the image of $\Phi: Mat_{n_1n_2}(k) \to A$ must be fully determined by what it does to those two subalgebras. I think that it is actually determined by what it does to diagonal elements but I couldn't prove it.
I know that diagonal elements form a basis for matrix algebra as module over itself but it doesn't help. Also my guess never uses that images of metrices from different subalgebras commute.
I think you can proceed in the following way: The map
$$Mat_{n_1}(k) \times Mat_{n_2}(k) \to Mat_{n_1n_2}(k): (A,B) \mapsto A \odot B$$
is $k$-bilinear. Here $A \odot B$ denotes the kronecker-product of matrices.
Consequently, the universal property of the tensor product gives a $k$-algebra morphism
$$f: Mat_{n_1}(k)\otimes Mat_{n_2}(k) \to Mat_{n_1n_2}(k)$$
where $f(A \otimes B) = A \odot B$.
Show that $f$ is an isomorphism of $k$-algebra's.