I am a little unsure about my proof. I wanted to ask if this proof is correct.I would appreciate any suggestions and improvements. Futhermore I dont yet fully understand why in (a) from the surjectivity of the $Tx$ follows $y_1'=y_2'$. I hope someone can shed some light on the matter. Thank you in advance!
Let $X,Y$ be normed spaces and $T \in L(X,Y)$ (linear and bounded operator)
(1) If $T$ is surjective then $T'$ is injective
(2) If $T'$ is surjective then $T$ is injective
Def. dual operator T'
for $T \in L(X,Y)$ we say $T':Y'\to X'$ with $T'y'(x)=y'(Tx)$ is its dual operator
(1) Let $T'y_1'=T'y_2'$ then it is $T'y_1'(x)=T'y_2'(x)~\forall x \in X$. So by definition of dual operator: $$y_1'(Tx)=y_2'(Tx)$$
and since $Tx$ is surjective we have (a) $$y_1'(y)=y_2'(y)~\forall y =Tx$$
so $y'_1=y'_2$
(2)
Let $Tx_1=Tx_2$ and $x' \in X'$ an arbitrary functional then by surjectivity there is a $y' \in Y'$ such that $T'y'=x'$
It is: $$y'(Tx_1)=y'(Tx_2)~~\textrm{(because y' is linear and } Tx_1=Tx_2)$$
and this is equivalent to
$$T'y'(x_1)=T'y'(x_2) \iff x'(x_1)=x'(x_2)~\forall x'\in X'$$
It follows $x_1=x_2$