Show that $T\to T^*$ is an isomorphism (Where T is a linear transform) I think I solved it, but I used a dirty trick, I'd like someone to review it, that would be great.
Let $X,Y$ be linear spaces over field $F$. and $T:X \to Y$ a linear transformation.For each $T$ we define $T^*:Y^* \to X^*$ such that $T^*(\phi)(x)=\phi(Tx)$ where $x \in X, \phi \in Y^*$
Show that $Star: Hom(X,Y)\to Hom(Y^*,X^*)$, $Star(T)=T^*$ is an isomorphism.
My solution (For showing its injective, I'll work on surjective after I confirm this is correct):
Let's assume $Star(T_1)=Star(T_2)$, meaning $T_1^* = T_2^*$, which implies:
$\phi(T_1x)=\phi(T_2x)$ for all $\phi, x$. Now here comes the dirty trick I used (which is maybe wrong), if it's true for all $\phi \in Y^*$, then it's true also for $\phi' \in Y^*$ such that $\phi'$ is injective.
So we get $\phi'(T_1x)=\phi'(T_2x)$ but we assumed $\phi'$ is injective, so it follows that $T_1x=T_2x$ for all values of $x$ and so $T_1=T_2$ and so $Star$ is injective.
Is this argument correct?
 A: Assuming $X, Y$ have finite dimension, I'll show that $*$ injects.
Assume that $T_1^* = T_2^*$. Then, for any $\phi \in Y^*$, we have that
$$
T^*_1 \phi = T^*_2 \phi
$$
which literally means that for any $x \in X$, we have
$$
T^*_1 \phi(x) = T^*_2 \phi(x)
$$
Peeling back the definition of the transpose $*$, we have that for any $x \in X, \phi \in Y^*$, we have the equality
$$
\phi(T_1 x) = \phi(T_2 x)
$$
Using the linearity of $\phi$, we can re-write this as $\phi(T_1 x - T_2 x) = 0$.
Let $\{y_1, \cdots, y_n\}$ be a basis for $Y$ and let $\{\phi_1, \cdots, \phi_n\}$ be a dual basis for $Y^*$, i.e. $\phi_i(y_j) = \delta_{ij}$ (the Kronecker delta). 
It is easy (check this!) that if $y \in Y$ and $\phi_i(y) = 0$ for each of $i = 1, 2, \cdots, n$, then $y = 0$. But we already have that $\phi_i(T_1 x - T_2 x)  = 0$ for each $i = 1,\cdots,n$, and so $T_1 x - T_2 x = 0$. This completes the proof.
If you use the canonical injection $J : Y \to Y^{**}$, then you can extend this proof to show that $*$ is an injection regardless of dimension. On the other hand, as has already been observed, $*$ is not a surjection in every case. However, since $Hom(X,Y)$ and $Hom(Y^*, X^*)$ are of the same finite dimension, having shown the injectivity of $*$, we get the surjectivity for free.
A: The map $T\mapsto T^*$ is linear, with respect to the obvious vector space structure of $\operatorname{Hom}(X,Y)$. Thus you only need to show that $T\ne0$ implies $T^*\ne0$.
Suppose you have $T\ne0$; there is $x\in X$ with $Tx\ne0$. Complete $Tx$ to a basis of $Y$ and define $\phi\colon Y\to F$ to be $1$ on $Tx$ and $0$ on all the other elements of the basis. Then $\phi(Tx)=1\ne0$ and so $T^*(\phi)(x)=1\ne0$. Thus $T^*(\phi)\ne0$ and, finally, $T^*\ne0$.
Note that you don't need that $X$ and $Y$ be finite dimensional (this is needed for surjectivity).
Your argument fails because nothing makes it possible to choose an injective $\phi'\in Y^*$. Actually such an injective linear map can exist only if $Y$ has dimension $0$ or $1$.
A: What you need to use is that $Y^*$ separates points in $Y$. This means that given any $y_1\in Y$, there exists $\phi\in Y^*$ with $\phi(y_1)=1$. This is trivial to construct using basis in arbitrary vector spaces, or the Hahn-Banach Theorem in normed vector spaces. 
So now when you have $\phi(T_1x)=\phi(T_2x)$ for all $\phi\in Y^*$, you get that $T_1x=T_2x$. 
