Property-like structure in a model category

Let us say that a map $f : X \to Y$ in an $\infty$-category is an embedding if, for every object $Z$, the map of spaces $\mathrm{Hom}(Z,X) \to \mathrm{Hom}(Z,Y)$ is an embedding (that is, has either empty or contractible fibers). It is easy to define this notion on the level of model categories. A map $f : X \to Y$ in a model category is an embedding if, for every (cofibrant) object $Z$, the map of homotopy function complexes $\mathrm{hMap}(Z,X) \to \mathrm{hMap}(Z,Y)$ is an embedding.

I believe this is all that can be said in a general model category about this notion. The reason is that embeddings in $\infty$-categories can behave very adly (just as monomorphisms in ordinary categories), so there is no reason to believe that there should be some nice general theory that allows to work with them in arbitrary model category.

If we have some specific model category in mind, then we can try to choose some nice presentation of homotopy function complexes to prove that some specific map is an embedding. For example, in the case of the Joyal model structure, there are at least two such choices. Both of them were discussed here: Homotopy function complex for quasi-categories.

Now, let's say we have a map of quasicategories $i : A \to B$ and we want to show that, for every quasicategory $X$, the map $X^B \to X^A$ is an embedding ($i$ is the inclusion of the free idempotent into the free retract in your example). This is equivalent to asking the map $\mathrm{hMap}(B,X) \to \mathrm{hMap}(A,X)$ to be an embedding. If we choose the first presentation of $\mathrm{hMap}$ in the question cited above, then this map is an embedding if and only if $X$ has the right lifting property with respect to the pushout-product of $i$ with maps $\partial \Delta'[n] \to \Delta'[n]$, $n > 0$ (we need to assume that $i$ is a cofibration for this). Here, $\Delta'[n]$ is the nerve of the groupoid $\{ 0 \simeq \ldots \simeq n \}$.

This is similar to the situation with the map $\Lambda^1[2] \to \Delta[2]$ that you mentioned, but the problem is that objects $\Delta'[n]$ are more complicated than the objects $\Delta[n]$ that we have in the latter, so it is quite hard to prove this directly. If we define $\mathrm{hMap}(X,Y)$ as the largest Kan complex contained in $Y^X$, then the combinatorics of the involved simplicial sets becomes more manageable, but even then I don't see how to prove this directly.

Specifically for the case of quasi-categories (or any other model for $\infty$-categories) the following observation can be useful: suppose that $f: {\cal C} \to {\cal D}$ is a map of quasi-categories such that the induced map ${\rm Tw}(f):{\rm Tw}({\cal C}) \to {\rm Tw}({\cal D})$ on twisted arrow categories is coinitial, i.e., for every $e \in {\rm Tw}({\cal D})$ the comma category ${\rm Tw}({\cal C})_{/e}$ is weakly contractible (this is the dual property of cofinal, and implies in particular that reindexing diagrams along ${\rm Tw}(f)$ induces an equivalence on limits, rather than colimits). Then for any quasi-category ${\cal E}$ the induced map on functor categories ${\rm Fun}({\cal D},{\cal E}) \to {\rm Fun}({\cal D},{\cal E})$ is fully-faithful (and in particular, the induced map on functor spaces is an embedding). To see this, use the fact that if $g,h: {\cal D} \to {\cal E}$ are two functors then the space of natural transformations from $g$ to $h$ can be written as a limit on the twisted arrow category: $$ {\rm Map}_{{\rm Fun}({\cal D},{\cal E})}(g,h) \simeq {\rm lim}_{x \to y \in {\rm Tw}({\cal D})} {\rm Map}_{\cal E}(g(x),h(y)) $$ The coinitiality of ${\rm Tw}(f)$ now implies that ${\rm Map}_{{\rm Fun}({\cal D},{\cal E})}(g,h) \stackrel{\simeq}{\to} {\rm Map}_{{\rm Fun}({\cal C},{\cal E})}(g \circ f,h \circ f)$ is an equivalence.

Examples:

1) The map $\Delta^1 \to J$ from the walking arrow to the walking isomorphism induces a coinitial map on twisted arrow categories (this is because ${\rm Tw}(J) \simeq J \simeq \ast$ and ${\rm Tw}(\Delta^1)$ has a contractible geometric realization). This reflect the (simple, but important) fact that an arrow being an equivalence is a property.

2) Consider the map ${\rm Idem} \to {\rm Ret}$ from the free idempotent to the free retract (so that ${\rm Idem}$ has a single object equipped with an idempotent self map and ${\rm Ret}$ has two objects which sit in a retract diagram). Then one can show that the map ${\rm Tw}({\rm Idem}) \to {\rm Tw}({\rm Ret})$ is coinitial (this is a bit more tedious, but can still be done completely by hand: ${\rm Tw}({\rm Idem})$ has two objects, ${\rm Tw}({\rm Ret})$ has five). This reflects the fact you mentioned that a splitting of an idempotent is a property.