A linear map from $\mathbb{C}^n \otimes \overline{\mathbb{C}^n}$ to $M_n(\mathbb{C})$ $\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}$
I am trying to establish some properties of the linear map: $T:\mathbb{C}^n \otimes \overline{\mathbb{C}^n} \to M_n(\mathbb{C})$ given by: $T(\ket{i} \otimes \ket{j}) = \ket{i}\bra{j} \text{ , } 0 \leq i,j \leq n-1$. 
Where $\otimes$ denotes the tensor product, $\overline{\mathbb{C}^n}$ is the conjugate space of the complex (finite) Hilbert space $\mathbb{C}^n$, and $M_n(\mathbb{C})$ is the set of $n\times n$ matrices over $\mathbb{C}$.
I want to show the following:

(i) T is a bijection that preserves inner products (under the
  Hilbert-Schmidt inner product on $M_n(\mathbb{C})$
(ii) $T(\ket{\psi}\otimes\ket{\varphi} = \ket{\psi}\bra{\varphi}$ for
  every $\ket{\psi}, \ket{\varphi} \in \mathbb{C}^n$.

I think part of my trouble is figuring out the Dirac notation / how the outer product works. I'm not sure exactly what $\ket{i}\bra{j}$ looks like as a matrix (obviously it's a matrix in $M_n(\mathbb{C})$ but I'm not quite sure what the entries look like).
Also, I think part (ii) of the question is an immediate consequence from the fact that $T$ is a linear, inner product preserving bijection, but I'm a bit confused about why $\ket{\varphi}$ is now coming from $\mathbb{C}^n$ rather than $\overline{\mathbb{C}^n}$.
Any help is appreciated.
 A: In my opinion, this is one of several troubles with Dirac's notation. When you see the elements of $\mathbb C^n$ ("kets") as columns vectors, the "bras" are the adjoints, i.e. their conjugate tranposes. So you have, in $\mathbb C^2$, 
$$
|1\rangle=\begin{bmatrix} 1\\0\end{bmatrix},\ \ \langle 1|=\begin{bmatrix} 1\\0\end{bmatrix}^*=\begin{bmatrix} 1&0\end{bmatrix}. 
$$
Mathematicians usually use $\{e_1,\ldots,e_n\}$ for the canonical basis of $\mathbb C^n$, and so your map is 
$$
T(e_i\otimes e_j)=e_ie_j^*. 
$$
And now you can calculate the entries of the matrix explicitly. 
To see that $T$ is a bijection, you just check that $\{e_i\otimes e_j\}_{i,j=1,\ldots,n}$ 
 is a  basis of $\mathbb C^n\otimes\mathbb C^n$, and that $\{e_ie_j^*\}_{i,j=1,\ldots,n}$ is a basis of $M_n(\mathbb C)$ (we usually call the matrices $e_ie_j^*$ the "matrix units"). 
For the inner product, since the aforementioned bases are orthonormal bases, it is enough to check that $T$ preserves the inner product of basis elements. We have 
\begin{align}
\langle T(e_i\otimes e_j),T(e_s\otimes e_t)\rangle
&=\langle e_ie_j^*,e_se_t^*\rangle=\operatorname{Tr}((e_se_t^*)^*e_ie_j^*)
=\operatorname{Tr}(e_te_s^*e_ie_j^*)
=e_s^*e_i\,\operatorname{Tr}(e_te_j^*)\\
&=\delta_{i,s}\,\delta_{j,t}=\langle e_i,e_s\rangle\,\langle e_j,e_t\rangle\\
&=\langle e_i\otimes e_j,e_s\otimes e_t\rangle.
\end{align}
As for your last question, $\langle \varphi|\in\overline{\mathbb C^n}$ is the adjoint of $|\varphi\rangle\in\mathbb C^n$. The confusion arises is that the domain of $T$ is $\mathbb C\otimes \mathbb C$. Then, on the right, you take the adjoint of the second coordinate. 
And yes, part (ii) follows as you say: if $\varphi=\sum_{s=1}^n \varphi_se_s$ and $\psi=\sum_{t=1}^n \psi_te_t$, then 
\begin{align}
T(\varphi\otimes\psi)
&=\sum_{s,t} \varphi_s\overline{\psi_t}\,T(e_s\otimes e_t)\\
&=\sum_{s,t} \varphi_s\overline{\psi_t}\,e_se_t^*\\
&=\left(\sum_s\varphi_se_s\right)\,\left(\sum_t\psi_te_t\right)^*\\
&=\varphi\psi^*. 
\end{align}
A: Since Martin Argerami already gave the answer, the purpose of my post is to clarify the Dirac notation.
Let $|\varphi\rangle$ denotes a vector in some complex inner product space $(V, \langle\cdot \mid \cdot\rangle)$ and $\langle \psi|$ denotes a dual vector in $V^\ast$. Here $ \langle \psi|$ maps $V$ to $\mathbb{C}$ via $\langle \psi|\Big(|\varphi\rangle \Big)=\langle \psi \mid \varphi\rangle \in \mathbb{C}$.
Matrix Representation of Vectors: 
If $\mathcal{B}=\{|j\rangle\}_{j=1}^n$ is an orthonormal basis for $V$ and $\mathcal{B}^\ast=\{\langle j|\}_{j=1}^n$ is the dual basis for $V^\ast$ (i.e. $\langle i\mid j\rangle = \delta_{i, j}$, the Kronecker delta), then we see that
\begin{align}
|\varphi\rangle = \sum^n_{j=1}\alpha_j |j\rangle=\sum^n_{j=1} \langle j\mid\varphi\rangle |j\rangle   \ \ \text{ and } \ \ \langle \varphi| = \sum^n_{j=1} \beta_j\langle j| = \sum^n_{j=1}\langle\varphi\mid j\rangle \langle j|
\end{align}
or in matrix form we have the array of complex numbers
\begin{align}
[|\varphi\rangle]_\mathcal{B} =
\begin{pmatrix}
\langle 1\mid \varphi\rangle\\
\langle 2 \mid \varphi\rangle\\
\vdots\\
\langle n\mid \varphi\rangle
\end{pmatrix}
\in \mathbb{C}^n \ \ \text{ and } \ \ [\langle \varphi|]_{\mathcal{B}^\ast} = 
\begin{pmatrix}
\langle  \varphi\mid 1\rangle, \langle  \varphi\mid 2\rangle, \ldots, \langle  \varphi\mid n\rangle
\end{pmatrix} \in (\mathbb{C}^n)^\ast.
\end{align}
By the conjugate symmetry of the complex inner product ($\langle \varphi\mid \psi\rangle = \overline{\langle \psi\mid \varphi\rangle}=: \langle \psi^\ast\mid \varphi^\ast\rangle$), we see that the two arrays of numbers are related to one another via conjugate transpose. Hence it follows
\begin{align}
\langle \psi \mid \varphi\rangle = \sum^n_{j=1} \langle \psi|j\rangle \langle j\mid\varphi\rangle  =  \sum^n_{j=1} \overline{\langle j\mid \psi\rangle} \langle j\mid\varphi\rangle. 
\end{align}
Matrix Representation of Some Operators: Next, by rewriting $|\varphi\rangle$, one can get
\begin{align}
|\varphi\rangle = \sum^n_{j=1}|j\rangle\langle j\mid \varphi\rangle  =: \left(\sum^n_{j=1}|j\rangle\langle j| \right)|\varphi\rangle
\end{align}
where $|j\rangle \langle j|$ are defined to be linear transformations from $V\rightarrow V$ via $|j\rangle \langle j|\Big(|\varphi\rangle \Big)=\langle j\mid \varphi\rangle|j\rangle$. Then it follows that
\begin{align}
\sum^n_{j=1}|j\rangle\langle j| = I
\end{align}
called the resolution of the identity. Also, note that matrix representation of $|j\rangle\langle j|$ in $\mathcal{B}$ is given by
\begin{align}
[|1\rangle\langle 1|]_\mathcal{B} =
\begin{pmatrix}
1\\
0 \\
\vdots\\
0
\end{pmatrix}
\begin{pmatrix}
1 & 0& \cdots & 0
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & \cdots & 0\\
0 & & & \\
\vdots & &\mathbf{0} & \\
0 & & & 
\end{pmatrix} \in M_n(\mathbb{C}).
\end{align}
Likewise for $j=2, \ldots, n$. Also, we called $|j\rangle \langle j|$ the projection onto $|j\rangle$. 
In the more general case of $|\varphi\rangle \langle \psi|:V\rightarrow V$, it's not hard to see that
\begin{align}
[|\varphi\rangle \langle \psi|]_\mathcal{B} =&\ 
\begin{pmatrix}
\langle 1\mid \varphi\rangle\\
\langle 2 \mid \varphi\rangle\\
\vdots\\
\langle n\mid \varphi\rangle
\end{pmatrix}
\begin{pmatrix}
\langle  \psi\mid 1\rangle & \langle  \psi\mid 2\rangle &  \ldots &\langle  \psi\mid n\rangle
\end{pmatrix}\\
=&
\begin{pmatrix}
\langle 1\mid \varphi\rangle \langle \psi\mid 1\rangle & \langle 1\mid \varphi\rangle\langle \psi\mid 2\rangle &  \cdots & \cdots &  \langle 1\mid \varphi\rangle \langle \psi\mid n\rangle \\
\langle 2\mid \varphi\rangle \langle \psi\mid 1\rangle & \langle 2\mid \varphi\rangle\langle \psi\mid 2\rangle &  \cdots & \cdots &  \langle 2\mid \varphi\rangle \langle \psi\mid n\rangle \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
\langle n-1\mid \varphi\rangle \langle \psi\mid 1\rangle & \cdots &  \cdots & \langle n-1\mid \varphi\rangle \langle \psi\mid n-1\rangle &  \langle n-1\mid \varphi\rangle \langle \psi\mid n\rangle \\
\langle n\mid \varphi\rangle \langle \psi\mid 1\rangle & \cdots &  \cdots & \langle n\mid \varphi\rangle \langle \psi\mid n\rangle &  \langle n\mid \varphi\rangle \langle \psi\mid n\rangle \\
\end{pmatrix}\in M_n(\mathbb{C})
\end{align}
where $\langle i\mid \varphi\rangle \langle \psi\mid j\rangle$ are called the matrix elements of $|\varphi\rangle\langle \psi|$. Moreover, we see that $\{|i\rangle \langle j|\}$ forms a basis for $M_n(\mathbb{C})$. 
Next, note that we can also write
\begin{align}
[|\varphi\rangle \langle \psi|]_\mathcal{B} =&\ 
\begin{pmatrix}
\langle 1\mid \varphi\rangle\\
\langle 2 \mid \varphi\rangle\\
\vdots\\
\langle n\mid \varphi\rangle
\end{pmatrix}
\begin{pmatrix}
\overline{\langle  1 \mid \psi\rangle} & \overline{\langle  2\mid \psi\rangle} &  \ldots & \overline{\langle  n\mid \psi\rangle}
\end{pmatrix}\\
 =&\ [|\varphi\rangle]_\mathcal{B}\otimes_o [|\psi^\ast\rangle]_\mathcal{B} =[|\varphi\rangle\otimes_o |\psi^\ast\rangle]_\mathcal{B}
\end{align}
where $\otimes_o$ indicates the outer product of two arrays. In short, we see that
\begin{align}
\mathbb{C}^n\otimes_0 (\mathbb{C}^n)^\ast=[V\otimes_o V^\ast]_\mathcal{B}=M_n(\mathbb{C})
\end{align} 
or equivalently in a more abstract language
\begin{align}
V\otimes V^\ast \simeq \text{Hom}(V, V).
\end{align}
Answer: 
As pointed out by Martin Argerami, what you want to show is that $V\otimes V \simeq V\otimes V^\ast$. Thus $T:\mathbb{C}^n\otimes \mathbb{C}^n\rightarrow \mathbb{C}^n\otimes (\mathbb{C}^n)^\ast$ via
\begin{align}
T\left( \begin{pmatrix}
\langle 1\mid \varphi\rangle\\
\langle 2 \mid \varphi\rangle\\
\vdots\\
\langle n\mid \varphi\rangle
\end{pmatrix}\otimes_o\begin{pmatrix}
\langle 1\mid \psi\rangle\\
\langle 2 \mid \psi\rangle\\
\vdots\\
\langle n\mid \psi\rangle
\end{pmatrix}\right)
=\begin{pmatrix}
\langle 1\mid \varphi\rangle\\
\langle 2 \mid \varphi\rangle\\
\vdots\\
\langle n\mid \varphi\rangle
\end{pmatrix}\otimes_o\begin{pmatrix}
\overline{\langle 1\mid \psi\rangle}\\
\overline{\langle 2 \mid \psi\rangle}\\
\vdots\\
\overline{\langle n\mid \psi\rangle}
\end{pmatrix}.
\end{align}
