Throughout, let the morphism $A \overset{f}{\to} B$ be given. We say that "the extension problem for $f$ along $A \overset{t}{\to} T$ has a solution" if there exists a morphism $B \overset{g}{\to} T$ such that $g \circ f = t$.
(We are not requiring $t$ to be a monomorphism, as is commonly done according to nLab.)
One can show that if $f$ is a split monomorphism, then for all objects $T$ in the category, $f$ solves the extension problem for all $A \overset{t}{\to} T$. The converse is also true, since a retraction is just a solution to the extension problem for $f$ along $A \overset{id_A}{\to}A$.
Question: Is there a name for a morphism $f$ such that, for a given object $T$ (i.e. not necessarily all objects $T$) in the category, the extension problem for $f$ along every $A \overset{t}{\to} T$ has a solution?
I don't think this is the same thing as $f$ "having the left lifting property with respect to $1_T$", since it seems like every morphism (trivially) "has the left lifting property with respect to $1_T$".
Also this has to be strictly stronger than having the left cancellative property for morphisms $T \to A$, since if it weren't then every monomorphism would be a split monomorphism.
Lawvere and Schanuel call the analogous property for lifts (instead of extensions) "surjective for maps from $A$", but what they call "injective for maps from T" corresponds to the strictly weaker property mentioned above. I.e. a morphism is a monomorphism (not necessarily split) if and only if it is "injective for maps from $T$" for all $T$, but a morphism is "surjective for maps from $T$" for all $T$ if and only if it is a split epimorphism.
So there seems to be no name for the "single $T$" property for either arbitrary epimorphisms, or for split monomorphisms -- my question is above is about the terminology for split monomorphisms.