17

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?

kaba
  • 2,973
  • 18
  • 20
  • 1
    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). – Arturo Magidin Jun 20 '19 at 21:00
  • 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"? – Daniel Schepler Jun 20 '19 at 21:07
  • (And incidentally, since the underlying set functor on the category of compact Hausdorff topological spaces is monadic, a corollary is that it reflects isomorphisms.) – Daniel Schepler Jun 20 '19 at 21:09
  • 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. – Daniel Schepler Jun 20 '19 at 21:13
  • Related: https://math.stackexchange.com/questions/1222237/a-characterization-of-categories-with-a-conservative-forgetful-functor-to-set – Kevin Carlson Jun 20 '19 at 23:50

3 Answers3

13

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.

Ittay Weiss
  • 82,314
  • 1
    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. – Daniel Schepler Jun 20 '19 at 21:24
7

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.

Ethan Bolker
  • 107,009
2

Elementary examples include: bijective homomorphism of (groups/rings/modules/vector spaces) are isomorphisms.

Some slightly more advanced examples include bijective homomorphism of vector bundles / Lie groups are also isomorphisms.

The canonical non-examples: a bijective continuous map needs not be a homeomorphism; a bijective smooth map needs not be a diffeomorphism; a homeomorphism of algebraic varieties needs not be an isomorphism.

Sardines
  • 1,085