Isomorphisms in $\mathbf{Top}$ vs $\mathbf{Ring}$. Does it have some consequences?

Question

I was wondering why isomorphisms in some ''algebraic'' categories like $\mathbf{Grp}$ or $\mathbf{Ring}$ are simply morphisms which are also a bijection while in other concrete categories such as $\mathbf{Top}$ or $\mathbf{Man}$ that is not enough. In particular, I would like to know if this fact has some consequences for the properties of such categories.

I am not an expert in category theory at all, I have studied only the main definitions by my own. But this fact called my attention and I thought that, if it means something, the best way to study it would be through the language of category theory.

So, do we know if $\mathbf{Top}$ and $\mathbf{Ring}$ exhibit a difference behavior in some aspect (e.g. existence of limits) as a consequence of that?

Answer 1

First, for terminology, we say that a functor $F : \mathcal C → \mathcal D$ reflects isomorphisms or that it's conservative if for every morphism $f$ of $\mathcal C$ we have that if $Ff$ is iso, then $f$ is itself iso. So to say that all bijections in a concrete category are isomorphisms is the same as saying that its forgetful functor is conservative.

Here is a couple of consequences of a this, although they are somewhat related.

Conservative functors reflect all limits they preserve (and that exist in the source category). Concretely, in your two examples ($\mathrm{Top}$ and $\mathrm{Ring}$) all limits exist and are preserved by the forgetful functors, but in $\mathrm{Top}$, there are in general many ways in which $X \xleftarrow{π_1} (X × Y, \mathcal T) \xrightarrow{π_2} Y$ can fail to be the (topological) product of $X$ and $Y$ (eg. by taking the discrete topology for $\mathcal T$). This cannot happen in algebraic categories; there the algebraic structure on $X × Y$ is determined completely by the requirement that the projection maps be homomorphisms.

Second, the forgetful functor $U : \mathcal C → \mathrm{Set}$ is often right adjoint, ie. there exists the free object functor $F : \mathrm{Set} → \mathcal C$. In this situation you always have a canonical epimorphism $ε_X : FUX → X$, so that every object $X$ of $\mathcal C$ is in a sense a quotient of a free object. It's just that the "in a sense" of the previous sentence works a lot better if $U$ is conservative, than when it isn't (nb. you need $\mathcal C$ to have equalizers for this to be true, but that is a very mild condition). Concretely, in $\mathrm{Top}$ you get the canonical epimorphism $(X, \mathcal PX) → (X, \mathcal T)$ which in general most certainly isn't a quotient map of topological spaces in the usual sense. In $\mathrm{Ring}$ on the other hand you get $ℤ[UR] → R$ (where $ℤ[UR]$ is the polynomial ring in $UR$ variables), which is the canonical generators-and-relations representation of $R$, and a quotient in every sense of the word.

(To be precise, if (and only if!) $U$ is conservative, $ε_X$ will be extremal epimorphisms, which means that $ε_X$ cannot be factored as $FUX \xrightarrow e X' \xrightarrow m X$, where $m$ is a monomorphism which isn't iso (in other words, $ε_X$ doesn't factor through a proper subobject of $X$). This is something you'd certainly want your epimorphisms to satisfy, but it isn't true in general in either $\mathrm{Top}$ or $\mathrm{Ring}$ (identity from a finer to coarser topology, which is exactly what we got above, is an example in $\mathrm{Top}$, and localizations, eg. $ℤ → ℚ$, in $\mathrm{Ring}$), so the fact that $ε_X$ is extremal in $\mathrm{Ring}$ is particularly meaningful.)

Finally and somewhat tangentially, it is amusing to note that the forgetful functor from $\mathrm{Top}$ almost trivially satisfies the conditions of Beck's monadicity theorem except for the conservativity (the key point however is failure to reflect certain coequalizers), and monadic categories correspond exactly to one kind of algebraic theories (potentionally infinitary, with equational axioms).

Answer 2

Categorically, you have to define "isomorphism" only in terms of what is available - objects, morphisms, composition, identity. Thus we define an isomorphism as a morphism $f\colon A\to B$ that has a two-sided inverse morphism (i.e., $g\colon B\to A$ with $f\circ g=\operatorname{id}_B$ and $g\circ f=\operatorname{id}_A$).

In friendly algebraic categories, objects are just sets with some additional structure in form of (binary) operations and morphisms are maps between these sets that respect the additional operations, e.g., whenever $x+_Ay=z$ holds in $A$ we demand that $f(x)+_Bf(y)=f(z)$ holds in $B$. If a morphism $f$, viewed as a map between sets, is bijective, we at least have a two-sided inverse map $g\colon B\to A$. This turns out to be a momorphism: If $u+_Bv=w$ holds in $B$, then let $x=g(u)$, $y=g(v)$, and $z=x+y$. Then $w=u+_Bv=f(x)+_Bf(y)=f(z)$ and hence $z=g(w)$. It follows that the inverse of a bijective morphism is a morphism, as desired.

For more general structures (e.g., topologies, orders, ...), not ot mention for cases where it makes no sense to consider morphisms as maps, the argument from the fpreceding paragraph does not work so nicely.

Answer 3

The point of category theory is that there are certain properties that are uniform across categories (associativity of composition; existence of identity maps), and other properties that are not uniform across categories.

One nonuniform property amongst "concrete" categories (i.e. categories which have a forgetful functor to the category of sets): it may or may not be true that a morphism which is invertible as a set map is also invertible as a morphism. In fact when you put it that way, it seems perhaps unreasonable to expect.

So when you specialize your attention to a particular category, be sure you know which way that property goes, as well as others.