Why is the ring homomorphism $\Gamma: D^{o}\rightarrow \text{End}_{M_n(D)}(D^n)$, where $D$ is a division ring, a bijection? It is said in this answer that for any division ring $D$, the following ring homomorphism is a bijection:
$\Gamma: D^{o}\rightarrow \text{End}_{M_n(D)}(D^n)=\{M_n(D)\text{-module endomorphisms of }D^n\}\\d\mapsto \theta_d\quad \text{s.t.}\quad \theta_d(x_1,\ldots,x_n)=(x_1d,\ldots,x_nd)$
I can see why it is an injective homomorphism (if $d\neq d’$, $(x_1 d,\ldots,x_n d) \neq (x_1 d’,\ldots,x_n d’)$) but I’m not sure about how to prove it is surjective too. How can I write every element of $\text{End}_{M_n(D)}(D^n)$ in that form?
Oh, I also wanted to know what does the notation $D^o$ exactly mean here.
 A: Here $D^o$ refers to the opposite ring of $D$, i.e. $D$ but with its order of multiplication reversed.  This is needed in order for $\Gamma$ to be a ring homomorphism, since $\theta_{de}(x)=\theta_e(\theta_d(x))$ with the order of $e$ and $d$ reversed (right multiplication by $de$ means you first multiply by $d$ and then by $e$), so $\Gamma$ only preserves multiplication if you reverse the order of multiplication on the domain or codomain.
Surjectivity of $\Gamma$ is indeed nontrivial.  To prove it, suppose $f:D^n\to D^n$ is an $M_n(D)$-module homomorphism. Let $e_i$ denote the $i$th standard basis vector of $D^n$, and let $E_{ij}\in M_n(D)$ denote the matrix whose $ij$ entry is $1$ and all other entries are $0$.  Note that $E_{ij}e_j=e_i$ and $E_{ij}e_k=0$ if $k\neq j$.
Now write $f(e_1)=\sum a_ie_i$ for some $a_i\in D$.  Note that for each $k\neq 1$, we have $E_{jk}e_1=0$ and hence $$0=f(E_{jk}e_1)=E_{jk}f(e_1)=\sum_i a_iE_{jk}e_i=a_ke_j.$$  Thus $a_k=0$ for all $k\neq 1$.  That is, $f(e_1)=de_1$ where $d=a_1$.  For each $i$, we then have $$f(e_i)=f(E_{i1}e_1)=E_{i1}f(e_1)=dE_{i1}e_1=de_i.$$
Finally, for arbitrary $x=(x_1,\dots,x_n)\in D^n$, we have $x=\sum x_ie_i$ so $$f(x)=\sum x_if(e_i)=\sum x_ide_i=\theta_d(x).$$
