Let the fidelity between two quantum states be defined as
$$F(\rho, \sigma) = \|\sqrt{\rho}\sqrt{\sigma}\|_1.$$
If $\rho = \vert\psi\rangle\langle\psi\vert$, then $F(\rho, \sigma) = \sqrt{\langle\psi\vert\sigma\vert\psi\rangle}$.
Let us look at a teleportation protocol of a Bell state, where the two classical bits are sent with an error at most $\varepsilon$. That is, the set of messages from $\{00,01,10,11\}$ is received correctly with probability at least $1-\varepsilon$ and incorrectly with probability at most $\varepsilon$.
If the state to be sent was the Bell state $\vert\psi\rangle\langle\psi\vert$, this implies that the teleported state is simply
$$\omega = (1-\varepsilon)\vert\psi\rangle\langle\psi\vert + \varepsilon\vert\psi^\perp\rangle\langle\psi^\perp\vert$$
The fidelity of the received state with respect to the intended state is
$$F(\omega, \vert\psi\rangle\langle\psi\vert) \geq \sqrt{1-\varepsilon}$$
Now, let's look at the reversed situation. Suppose one has a state transmission protocol that works with error at most $\sqrt{1-\varepsilon}$ in fidelity. That is, given a Bell state $\vert\psi\rangle\langle\psi\vert$, the state transmission protocol outputs a state $\omega$ such that
$$F(\omega, \vert\psi\rangle\langle\psi\vert) \geq \sqrt{1-\varepsilon}.$$
If one uses this protocol for superdense coding, what is the worst case error in decoding a message in the set $\{00,01,10,11\}$? I am not sure how to determine the worst case error post-measurement from the fidelity. Using Fuchs van de Graaf, the trace distance error is $\sqrt{\varepsilon}$ and therefore my post-measurement outcome has error $\sqrt{\varepsilon}$. But if teleportation and superdense coding were "reversible", it should be $\varepsilon$ here, shouldn't it?
So what is the error of correctly decoding messages in a superdense coding protocol if the Bell state is transmitted with fidelity of $\sqrt{1-\varepsilon}$? Did I "lose too much" in my analysis above to obtain the looser bound of $\sqrt{\varepsilon}$?