Adjoint of a linear isomorphic functional is an isomorphism I have this exercise to solve:

Let $E$ and $F$ be normed spaces and let $T \in L(E,F)$, where $L(E,F)$ denotes the set of all bounded linear operators from $E$ to $F$. $T^*$ is the dual operator of $T$ with $T^*: F^* \to E^*$ and $T^*\varphi=\varphi \circ T$. Assume $T$ is an isomorphism, i.e. $T$ is bijective and $T^{-1}\in L(F,E)$. Prove that $T^*$ is also an isomorphism and $(T^*)^{-1}=(T^{-1})^*$.

My work

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*Proof of $(T^*)^{-1}=(T^{-1})^*$:

We know that $(T \circ G)^*= T^*\circ G^*$ since $((T \circ G)^*\varphi)(x)=\varphi((T\circ G)(x))=(G^*\varphi)(Tx)=((T^*\circ G^*)\varphi)(x)$
Now plug in $T^*$ and $(T^{-1})^*$ to the equation, we have:
$T^*\circ (T^{-1})^* = (T\circ T^{-1})^*= Id^* = Id$ and $(T^{-1})^* \circ T^* = (T^{-1}\circ T)^*= Id^* = Id$
Hence, $(T^*)^{-1}=(T^{-1})^*$.

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*Proof of isomorphism:
Injectivity:
Let $\varphi$ and $\psi \in F^*$. Let $T^*\varphi=T^*\psi$ then we have $\varphi \circ T=\psi \circ T$. Here can I conclude that since $T$ bijective we have $\varphi = \psi$?
It remains to prove the surjectivity and I'm still struggling with this part.
 A: Hong thank you for tagging me in the other question: Let $X$ and $Y$ be Banach spaces, show that if they are isomorphic, then $X$ is reflexive iff $Y$ is reflexive.
But the answer is just that $(f^{-1})^*=(f^*)^{-1}$ and $f^*$ linear.
Do you need to see either of these facts spelled out further?  The point is that $f^*$ is an isomorphism if and only if $(f^*)^{-1}$ is continuous (or more generally, $g:X\to Y$ is an isomorphism if and only if $g^{-1}$ is continuous). But continuity is equivalent to being bounded.  So we just have to show that $(f^{-1})^*$ is bounded.  Does that clarify things?
See this for example:
http://mathonline.wikidot.com/isomorphism-linear-operators-on-normed-linear-spaces
A: Hint for surjectivity:
Let $\psi \in E^*$ be arbitrary.  You need to find a functional $\varphi \in F^*$ such that $T^* \varphi = \psi$.  That is, such that $\varphi \circ T = \psi$.  Can you guess how to define a functional $\varphi$ that might work?  You'll have to verify that your guess works, and that it's actually a continuous functional.
