For some "structures" (in informal sense for a lack of a formal term) in mathematics, such as groups, rings, and vector spaces, a bijective homomorphism is an isomorphism; i.e. the inverse is also a homomorphism. For some other structures, such as topological spaces and differentiable manifolds, a bijective homomorphism may not be an isomorphism.

Are there characterizations of sub-classes of structures which have the property that a bijective homomorphism is an isomorphism? For example, do all algebraic structures (in formal sense this time) have this property?

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    $\begingroup$ In categories of algebras yes, a morphism that is bijective on underlying sets is always an isomorphism (invertible morphism). You want to restrict to algebras-and-operations, as relational algebras and partial algebras do not have this property (since the partial operation may be defined on larger domains in the target). $\endgroup$ – Arturo Magidin Jun 20 at 21:00
  • $\begingroup$ In the study of monads, one will frequently see the definition that a functor $F : \mathbf{C} \to \mathbf{D}$ reflects isomorphisms if whenever $f$ is a morphism in $\mathbf{C}$ and $F(f)$ is an isomorphism (in $\mathbf{D}$) then that implies $f$ is an isomorphism. So, your question could be restated as "when does the underlying set functor reflect isomorphisms"? $\endgroup$ – Daniel Schepler Jun 20 at 21:07
  • $\begingroup$ (And incidentally, since the underlying set functor on the category of compact Hausdorff topological spaces is monadic, a corollary is that it reflects isomorphisms.) $\endgroup$ – Daniel Schepler Jun 20 at 21:09
  • $\begingroup$ Also somewhat tangentially related: you've probably seen the fact that if $F,G : \mathbf{C} \to \mathbf{D}$ are two functors and $\mu : F \to G$ a natural transformation such that $\mu_X$ is an isomorphism for each object $X$ of $\mathbb{C}$, then $\mu$ is an isomorphism of functors. $\endgroup$ – Daniel Schepler Jun 20 at 21:13
  • $\begingroup$ Related: math.stackexchange.com/questions/1222237/… $\endgroup$ – Kevin Carlson Jun 20 at 23:50

Recall that structures and their structure preserving maps often assemble themselves into categories. So, there is a category $Grp$ of groups, $Ab$ of abelian groups, $Ring$ of rings and so on. Now, if the structures are based on sets, then often there will be a forgetful functor $C\to Set$ from the category $C$ to the category of sets and functions. The property you are looking at is reflection of isomorphisms by this functor. So, one can ask, for a given category $C$, when is the forgetful functor $C\to Set$, assuming it exists, reflects isomorphism? A pretty far reaching answer is that whenever $C\to Set$ is monadic. Now, that latter term is a bit more technical, but, in a nutshell, $C\to Set$ is monadic if $C$ is a category of nice enough algebraic structures. Monadicity captures many algebraic structures, but not, for instance, posets (if you consider these algebraic): the forgetful functor does not reflect isomorphisms.

Interestingly, this notion of reflection of isomorphisms is in fact one of the conditions of Beck's Monadicity Theorem characterising monadic adjunctions.

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    $\begingroup$ As an interesting example, taking the canonical example of a common "algebraic structure" which is not a variety of algebras: The underlying set functor on the category of fields isn't monadic (for instance it doesn't create limits) though it does reflect isomorphisms. $\endgroup$ – Daniel Schepler Jun 20 at 21:24

A relevant anecdote but not an answer.

When I took abstract algebra (in 1956, way before categories) a question on the first exam asked for the definition of an isomorphism for an equivalence relation - not something we'd covered in class.

I (and most of my classmates) naively modified the definition we knew for group homomorphisms, requiring that the bijection $\phi$ satisfy $\phi(x) \equiv \phi(y)$ whenever $x \equiv y$. We all lost points.


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