How to prove injectivity of the map $A^N \otimes_A A^N \to M_N(A)$, which is supposed to be an isomorphism of $A$-bimodules? I'd like to show that for a (unital) algebra $A$, which may be infinite-dimensional, we have $A^N \otimes_A A^N \cong M_N(A)$ as $M_N(A)$-bimodules. Here the left (right) action of $M_N(A)$ on $M_N(A)$ is given by left (right) matrix multiplication, and the left and right action of $M_N(A)$ on $A^N\otimes_A A^N$ is given by $M(a\otimes b) =(Ma)\otimes b$ and $(a\otimes b)M = a\otimes (bM)$, respectively, where $Ma$ and $bM$ are again just left and right matrix multiplication, respectively. In other words: we use the canonical actions on both spaces.
It seems like the following map $A^N \otimes_A A^N \to M_N(A)$ should be an isomorphism of $M_N(A)$-bimodules. For $(a_1,\dots,a_N)\in A^N$ and $(b_1,\dots b_N)\in A^N$ define
\begin{align}
(a_1,\dots,a_N)\otimes (b_1,\dots b_N) \mapsto \begin{pmatrix}
a_1b_1 &\dots &a_1b_n \\
\vdots &\ddots & \vdots \\
a_Nb_1 &\dots & a_Nb_N
\end{pmatrix}
\end{align}
I've shown that the map is indeed well-defined, is linear, respects the $M_N(A)$ actions, and is surjective, but I don't quite see how to prove injectivity, so I'm hoping somebody can help me with this. 
I do see a resemblance with endomorphisms of finite-dimensional vector spaces, which are injective if and only if they are surjective, so in that case the proof would be complete. In my case, however, $A$ can be infinite-dimensional, so the domain and co-domain of the map can be as well, and the only option then (concerning this resemblance) is to view $A^N \otimes_A A^N$ and $M_N(A)$ as modules over the ring $A$ (note that $A$ is a ring, as it is a unital algebra), but I'm not sure if and how to the 'injective $\Leftrightarrow$ surjective - property' of vector spaces translates to modules over a ring. So maybe this path is not possible at all. (EDIT: As pointed out in the comments, it is not the case that injective implies surjective in the case of modules.)
Lastly, I'm wondering if the assumption that $A$ be unital is necessary, or if it is possible to prove that  $A^N \otimes_A A^N \cong M_N(A)$ without this assumption. I needed the unit for proving surjectivity, and in the proof for injectivity as proposed above as well.
Thanks!
 A: I found a really simple proof of bijectivity of the given map. Denote by $e_{ij}$ the matrix in $M_N(A)$ with the unit $e\in A$ in entry $ij$ and zeros everywhere else. Similarly, denote by $e_i$ the element in $A^N$ which has the unit $e$ in the $i$-th entry and zeros everywhere else. Obviously, under the proposed map, we have $e_i\otimes e_j\mapsto e_{ij}$ and hence $e_{ij}\mapsto e_i\otimes e_j$ is the inverse map. 
To see that these are indeed inverse maps we must note that every $(a_1,\dots a_n)\in A^N$ is of the form $(a_1,\dots a_n)=\sum a_ie_i$ and hence $(a_1,\dots a_n)\otimes (b_1,\dots b_n) = \sum a_ib_je_i\otimes e_j$, so that every element in $A^N\otimes_A A^N$ is of the form $\sum c_{ij}e_i\otimes e_j$ with $c_{ij}\in A$. Also, each element in $M_N(A)$ is of the form $\sum c_{ij}e_{ij}$. Under composition of maps we thus obtain
\begin{align}
\sum c_{ij}e_i\otimes e_j \mapsto \sum c_{ij}e_{ij}\mapsto \sum c_{ij}e_i\otimes e_j, \\
%
\sum c_{ij}e_{ij}\mapsto \sum c_{ij}e_i\otimes e_j \mapsto \sum c_{ij}e_{ij},
\end{align}
showing that the maps are each others inverse. Hence the map proposed in the question is bijective, and an $M_N(A)$-bimodule isomorphism.
