[proofplan]
We show that the canonical embedding $J_{X/Y}: X/Y \to (X/Y)^{**}$ is surjective. The key tool is the double [adjoint](/page/The%20Adjoint%20of%20an%20Operator) $\pi^{**}: X^{**} \to (X/Y)^{**}$ of the quotient map $\pi: X \to X/Y$. We first establish the commutative identity $\pi^{**} \circ J_X = J_{X/Y} \circ \pi$, then use [reflexivity](/page/Reflexive%20Space) of $X$ (surjectivity of $J_X$) and surjectivity of $\pi$ to conclude that $J_{X/Y}$ is surjective.
[/proofplan]
[step:Establish the commutative identity $\pi^{**} \circ J_X = J_{X/Y} \circ \pi$]
Let $\pi: X \to X/Y$ denote the quotient map, $\pi(x) = x + Y$. Its adjoint is the linear isometry
\begin{align*}
\pi^*: (X/Y)^* &\to X^* \\
h &\mapsto h \circ \pi,
\end{align*}
and the double adjoint is
\begin{align*}
\pi^{**}: X^{**} &\to (X/Y)^{**} \\
\xi &\mapsto \xi \circ \pi^*.
\end{align*}
For every $x \in X$ and every $h \in (X/Y)^*$, compute
\begin{align*}
(\pi^{**} \circ J_X)(x)(h) &= \pi^{**}(J_X(x))(h) = J_X(x)(\pi^*(h)) = J_X(x)(h \circ \pi) = (h \circ \pi)(x) = h(\pi(x)),
\end{align*}
and
\begin{align*}
(J_{X/Y} \circ \pi)(x)(h) &= J_{X/Y}(\pi(x))(h) = h(\pi(x)).
\end{align*}
Since both expressions equal $h(\pi(x))$ for every $x \in X$ and $h \in (X/Y)^*$, we conclude
\begin{align*}
\pi^{**} \circ J_X = J_{X/Y} \circ \pi.
\end{align*}
[guided]
The natural approach to reflexivity of quotients is to exploit the functoriality of the double dual. The quotient map $\pi: X \to X/Y$ induces, by dualising twice, a map $\pi^{**}: X^{**} \to (X/Y)^{**}$. These four maps form a commutative square: the two paths from $X$ to $(X/Y)^{**}$ — namely $\pi^{**} \circ J_X$ (embed into the bidual then apply $\pi^{**}$) and $J_{X/Y} \circ \pi$ (quotient first then embed into the bidual) — agree.
The verification is a direct computation: both paths send $x \in X$ to the functional $h \mapsto h(\pi(x))$ in $(X/Y)^{**}$.
The commutativity $\pi^{**} \circ J_X = J_{X/Y} \circ \pi$ holds for *any* bounded linear map between [normed spaces](/page/Normed%20Vector%20Space), not just quotient maps. It is a general property of canonical embeddings and double adjoints.
[/guided]
[/step]
[step:Conclude surjectivity of $J_{X/Y}$ from reflexivity of $X$]
We compute the image of $J_{X/Y}$. Since $X$ is reflexive, the canonical embedding $J_X: X \to X^{**}$ is surjective (by the [Canonical Embedding into the Bidual is an Isometry](/theorems/875), $J_X$ is an isometry, and reflexivity means it is also surjective). Therefore $J_X(X) = X^{**}$. Since $\pi: X \to X/Y$ is the quotient map, it is surjective: $\pi(X) = X/Y$.
Applying the commutative identity to determine the range of $\pi^{**}$:
\begin{align*}
\pi^{**}(X^{**}) = \pi^{**}(J_X(X)) = (J_{X/Y} \circ \pi)(X) = J_{X/Y}(\pi(X)) = J_{X/Y}(X/Y).
\end{align*}
It remains to show that $\pi^{**}$ is surjective, i.e., $\pi^{**}(X^{**}) = (X/Y)^{**}$.
[guided]
We have established that $\pi^{**}(X^{**}) = J_{X/Y}(X/Y)$. So the surjectivity of $J_{X/Y}$ (which is what reflexivity of $X/Y$ means) is equivalent to the surjectivity of $\pi^{**}$. We prove the latter in the next step.
[/guided]
[/step]
[step:Prove $\pi^{**}$ is surjective using the Hahn-Banach extension]
Let $\Phi \in (X/Y)^{**}$. We construct $\xi \in X^{**}$ with $\pi^{**}(\xi) = \Phi$.
The adjoint $\pi^*: (X/Y)^* \to X^*$ is a linear isometry whose image is the annihilator $Y^\perp := \{f \in X^* : f(y) = 0 \text{ for all } y \in Y\}$. This is the content of the [Dual of a Quotient Space](/theorems/978) isomorphism: $h \mapsto h \circ \pi$ identifies $(X/Y)^*$ isometrically with $Y^\perp$, and $Y^\perp$ is a closed subspace of $X^*$ (as the intersection of kernels of the [continuous](/page/Continuity) evaluation maps $f \mapsto f(y)$ for $y \in Y$).
Define the bounded linear functional
\begin{align*}
\Lambda: Y^\perp &\to \mathbb{R} \\
f &\mapsto \Phi((\pi^*)^{-1}(f)).
\end{align*}
This is well-defined since $\pi^*: (X/Y)^* \to Y^\perp$ is bijective. The map $\Lambda$ is linear (as the composition of two [linear maps](/page/Linear%20Map)) and bounded: for $f \in Y^\perp$, writing $h = (\pi^*)^{-1}(f)$ so that $f = h \circ \pi$ and $\|h\|_{(X/Y)^*} = \|f\|_{X^*}$ (since $\pi^*$ is an isometry),
\begin{align*}
|\Lambda(f)| = |\Phi(h)| \le \|\Phi\|_{(X/Y)^{**}} \|h\|_{(X/Y)^*} = \|\Phi\|_{(X/Y)^{**}} \|f\|_{X^*}.
\end{align*}
Therefore $\|\Lambda\|_{(Y^\perp)^*} \le \|\Phi\|_{(X/Y)^{**}}$.
By the [Norm-Preserving Extension of Linear Functionals](/theorems/880) applied to the normed space $X^*$ and its closed subspace $Y^\perp$, there exists $\xi \in (X^*)^* = X^{**}$ extending $\Lambda$:
\begin{align*}
\xi|_{Y^\perp} = \Lambda, \quad \|\xi\|_{X^{**}} = \|\Lambda\|_{(Y^\perp)^*}.
\end{align*}
We verify $\pi^{**}(\xi) = \Phi$. For any $h \in (X/Y)^*$:
\begin{align*}
\pi^{**}(\xi)(h) = \xi(\pi^*(h)) = \xi(h \circ \pi) = \Lambda(h \circ \pi) = \Phi((\pi^*)^{-1}(h \circ \pi)) = \Phi(h),
\end{align*}
where the third equality holds because $h \circ \pi = \pi^*(h) \in Y^\perp$ and $\xi$ agrees with $\Lambda$ on $Y^\perp$, and the fourth uses $(\pi^*)^{-1}(\pi^*(h)) = h$.
Therefore $\pi^{**}(\xi) = \Phi$, proving surjectivity of $\pi^{**}$.
[guided]
The problem is that $\pi^{**}(\xi) = \xi \circ \pi^*$, and $\pi^*$ maps $(X/Y)^*$ into $X^*$. So $\pi^{**}(\xi)$ only "sees" $\xi$ on the subspace $Y^\perp = \operatorname{Im}(\pi^*) \subset X^*$. To hit a given $\Phi \in (X/Y)^{**}$, we only need $\xi$ to have the right values on $Y^\perp$ — its values on the complement of $Y^\perp$ are irrelevant.
This is why Hahn-Banach is the right tool: we define $\xi$ on the closed subspace $Y^\perp$ (where its values are forced by $\Phi$) and extend arbitrarily to all of $X^*$.
The functional $\Lambda: Y^\perp \to \mathbb{R}$ is defined by transporting $\Phi$ through the isomorphism $\pi^*: (X/Y)^* \to Y^\perp$. Since $\pi^*$ is an isometry, $\Lambda$ inherits the bound $\|\Lambda\| \le \|\Phi\|$.
After applying the [Norm-Preserving Extension](/theorems/880) to obtain $\xi \in X^{**}$, the verification is straightforward: for any $h \in (X/Y)^*$, the vector $\pi^*(h) = h \circ \pi$ belongs to $Y^\perp$, so $\xi(\pi^*(h)) = \Lambda(\pi^*(h)) = \Phi(h)$.
[/guided]
[/step]
[step:Combine surjectivity of $\pi^{**}$ with the commutative identity to conclude $X/Y$ is reflexive]
From the previous steps:
\begin{align*}
(X/Y)^{**} = \pi^{**}(X^{**}) = J_{X/Y}(X/Y).
\end{align*}
The first equality is the surjectivity of $\pi^{**}$ (proven in the preceding step). The second equality is the commutative identity $\pi^{**}(X^{**}) = \pi^{**}(J_X(X)) = J_{X/Y}(\pi(X)) = J_{X/Y}(X/Y)$, using $J_X(X) = X^{**}$ (reflexivity of $X$) and $\pi(X) = X/Y$ (surjectivity of $\pi$).
Since $J_{X/Y}(X/Y) = (X/Y)^{**}$, the canonical embedding $J_{X/Y}$ is surjective, so $X/Y$ is reflexive.
[/step]
