Let $T$ be a first-order theory, and let $C$ be the category of models of $T$, viewed as a concrete category. Then my question is, is the forgetful functor of $C$ an isofibration?
The reason I ask is, this answer says that the forgetful functor of the category of algebras over a given monad is an isofibration. But the category of fields is not a category of algebras over a monad, yet its forgetful functor is still an isofibration. So I’d like to find out if that’s just a coincidence or something true of all categories of models of a theory.
Yes, of course. If I understand the definition of isofibration correctly, this unpacks to the standard "transfer of structure". Given a model $M$ and a bijection $f$ from the underlying set of $M$ to another set $X$, we can equip $X$ with a structure $N$ in the obvious way, which makes $f$ into an isomorphism from $M$ to $N$.
This has nothing to do with $M$ being a model of a first-order theory. For any language $L$, the category $\mathsf{Str}_L$ of $L$-structures is a concrete category such that the forgetful functor to Set is an isofibration. Now if you take any subcategory of $\mathsf{Str}_L$ which contains all the isomorphisms in $\mathsf{Str}_L$, the restriction of the forgetful functor to this subcategory will still be an isofibration. So the same result holds for any subcategory of $\mathsf{Str}_L$ which contains all the $L$-isomorphisms. In particular, the category of models of any theory in any reasonable logic (which talks about $L$-structures).