n-k$. If $j \equiv (n-k) \pmod 2$, then using the fact that $j \geq k$, $$|{\psi}(x)| = \left| c_\frac{n-k+j}{2}  \circ 1 \right| = \frac{n-k+j}{2} + 1 > \frac{n}{2}.$$ Otherwise, $j \not\equiv (n-k) \pmod 2$. Since $n$ is even, it follows that $j \not \equiv k \pmod2$, and thus $j \geq k+1$. Hence, $$|{\psi}(x)| = \left| c_\frac{n-k+j+1}{2} \circ 0 \right| > \frac{n}{2}.$$ In both cases ${\psi}(x) \in {\mathcal{B}_{n}}$.

We conclude the proof by describing the inverse mapping ${\psi}^{-1} \colon {\mathcal{B}_{n}}\to {\{0,1\}^{n}}$. For $z \in {\mathcal{B}_{n}}$, write $z = x \circ z_{n+1}$, where $x \in {\{0,1\}^{n}}$ and $z_{n+1}$ is the $(n+1)^{\text{st}}$ bit of $z$. Let $C = \{ c_k, c_{k+1}, \ldots, c_{n-k}\}$ be the symmetric chain that contains $x$, and let $j$ be the index such that $x = c_j$ (note that $j \ge n/2$). Then, $$\label{eq:f inverse}
    {\psi}^{-1}(z) = \begin{cases}
                    c_{2j-(n-k)}    & \text{ if } z_{n+1} = 1; \\
                    c_{2j-(n-k)-1}    & \text{ if } z_{n+1} = 0.
                \end{cases}$$ It is straightforward to verify that this is indeed the inverse mapping of ${\psi}$.

In order to understand the mapping ${\psi}$ better, consider $x \in {\{0,1\}^{n}}$ and let $y \in {\{0,1,{\sqcup}\}^{n}}$ be the encoding of the chain that contains $x$. Note that if 