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Let $\rho_{AB}$ be a state and $T: B \rightarrow C$ be a CPTP map with $\sigma_{AC}= T(\rho_{AB})$. It is well known that $H_{\infty}(A \vert B)_{\rho} \geq H_{\infty}(A \vert C)_{\sigma}$ (aka data processing.) Furthermore, equality holds when $T=V$ is an isometry giving $H_{\infty}(A \vert B)_{\rho} = H_{\infty}(A \vert C)_{V\rho V^{\dagger}}$.

My question is, does the result hold if we consider $V^{\dagger}$ instead of $V$, that is, is the following equality true?

$$H_{\infty}(A \vert C)_{\sigma} = H_{\infty}(A \vert B)_{V^\dagger \sigma V}$$.

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No, the co-isometry map $\sigma \to V^\dagger \sigma V$ is not trace preserving. In the worst case you can have something like this. Take an isometry $V: \mathbb{C}^2 \rightarrow \mathbb{C}^3$ which just embeds $\mathbb{C}^2$ into $\mathbb{C}^3$ in the standard way i.e., $$ V = |0\rangle \langle 0 | + |1 \rangle \langle 1|. $$ This is an isometry $V^\dagger V = \mathbb{I}_{\mathbb{C}^2}$. But imagine now that we have a state $\sigma = |3 \rangle \langle 3|$ on $\mathbb{C}^3$, then $$ V^\dagger \sigma V = 0. $$ Which would result in an infinite min-entropy.

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