There is an inclusion $f : \mathbb{Z} \hookrightarrow \mathbb{R}$. Furthermore, if $A \subseteq \mathbb{R}$ has a maximum element, then so too does $f^{-1}(A)$. There is also an inclusion $g : \mathbb{Q} \hookrightarrow \mathbb{R}$. However, $g^{-1}(-\infty,\pi]$ does not have a maximum element, despite that $(-\infty,\pi]$ does.
Question. Suppose $Q$ and $R$ are linearly ordered sets, and that $f : Q \rightarrow R$ is an injective order-preserving map that is both cofinal and coinitial. If every subset of $R$ with a maximum element has the property that its preimage under $f$ has a maximum element, does it follow that every subset of $R$ with a minimum element has the property that its preimage under $f$ has a minimum element?
Please comment if any of the terminology needs clarification.
No. Let $R=\Bbb R$ and $Q=\left\{\frac1n:n\in\Bbb N\right\}\cup(\Bbb Z\setminus\{0\})$, and let $f$ be the inclusion map. If $S\subseteq R$ has a maximum element, so does $f^{-1}[S]$, but $f^{-1}\big[[0,1]\big]$ has no minimum element, even though $\min[0,1]=0$.