A Space is Reflexive if its Image under the Canonical Injection is Reflexive? Consider the following corollary in Brezis:

Here is part of the proof of it:

It was mentioned that if $J(E) \subseteq E^{**}$ is reflexive, then $E$ is also reflexive, where $J: E \to E^{**}$ be the canonical injection. Why is this true?
 A: This reduces to prove the following claim: Let $E$ and $F$ be Banach spaces. If $T:E \to F$ is a surjective isometry, then $E$ is reflexive if and only if $F$ is reflexive. We show only one direction as the other way is symmetric. Since $T$ is surjective and an isometry, it must be bijective as isometry implies injectivity. This implies that the adjoint operator $T^*: F^* \to E^*$ is also bijective as $(T^*)^{-1} = (T^{-1})^*$. Similarly we have $T^{**}: E^{**} \to F^{**}$ to be a bijection as well. Now we claim that the canonical injection
$$
J_F: F \to F^{**}
$$
can be written as $J_F \equiv T^{**} \circ J_E \circ T^{-1}$, where $J_E: E \to E^{**}$ is the canonical injection (in this case is bijective as $E$ is reflexive). Granted this is true, then $J_F$ is a bijection, so indeed a surjective mapping. Now to show the claim, we need to show
$$
T^{**} \circ J_E \circ T^{-1}(f) = J_F(f)
$$
for all $f \in F$. To show this, fix $f \in F$, we need to show
$$
T^{**} \circ J_E \circ T^{-1}(f)(x) = J_F(f)(x)
$$
for all $x \in F^{*}$. Now fix $x \in F^{*}$, we have
\begin{align}
T^{**} \circ J_E \circ T^{-1}(f)(x) &= (T^{**} \circ J_E \circ T^{-1}(f), x)\\
&= (J_E \circ T^{-1}(f), T^*x)\\
&= (T^*x, T^{-1}(f)) \\
&= (x, T T^{-1}(f))\\
&= (x, f) = J_F(f)(x).
\end{align}
This shows that we indeed have $J_F \equiv T^{**} \circ J_E \circ T^{-1}$.
