When does a map induce a structure in a concrete category?

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Let $C$ be a concrete category, let $A$ be a set, let $B$ be an object in $C$, and let $f$ be a function from $A$ to the underlying set of $B$. Then does there always exist an object in $C$ whose underlying set is $A$ such that $f$ is a morphism in $C$? And if $f$ is a bijection, then does there always exist an object in $C$ whose underlying set is $A$ such that $f$ is an isomorphism in $C$?

I assume the answer to these questions is no, but is there a name for concrete categories for which the answer to one or both questions is yes? What if we switched the order of $A$ and $B$, so that the set which is being turned into an object is the codomain of the function rather than the domain?

I ask because inducing a structure on a set via a map is a very common construction in mathematics, and I’m wondering whether it’s category-theoretic in origin.

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No. Just consider the category $1$ with one object and one morphism (the identity), and map that into Set in pretty much any way.

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On

You're basically asking about lifting properties of the concretising functor $U: C \rightarrow \underline{Set}$. This should give you an idea what to search for if you want to look up more references on this. Note that we can also study these properties when the target category of the functor is something other than $\underline{Set}$.

Functors where we can lift any morphism are very rare. Indeed, you generally wouldn't expect forgetful functors of a concrete category to have this property, simply because the idea of a concrete category is that most functions in $\underline{Set}$ aren't morphisms of the structures than make up the concrete category.

The case where we can only lift bijections (only lift isomorphisms, for a general target category) are well known and studied, with the functors in question being called isofibrations. Your intuition that there are many such cases comes from, among other things, the fact that the forgetful functor from the category of algebras of a monad is always an isofibration.