## Category Theory: Mono, Epi, but not Iso?

### January 2, 2013

This post will require some very basic knowledge of category theory (like, what a category is, and how to make a poset into a category).  For everything below, I will be a bit informal, but I will essentially mean that $A, B$ are objects in a category, and $f:A\to B$ is some morphism between them which is also in the category.

The "natural" extension of the notion of a surjective map (in, say, the category of sets) is

Definition.  A map $f:A\to B$ is an epimorphism if, for each object $Z$ and map $g,g':B\to Z$ we have that if $g\circ f = g'\circ f$ then $g = g'$.

You should prove for yourself that this is, in fact, what a surjective map "does" in the category of sets.  Pretty neat.  Similarly, for injective maps (in, say, the category of sets) we have the more general notion:

Definition. A map $f:A\to B$ is a monomorphism if, for each object $Z$ and map $g,g':Z\to A$ we have that if $f\circ g = f\circ g'$ then $g = g'$.

Again, you should prove for yourself that this is the property that injective mappings have in the category of sets.  Double neat.  There is also a relatively nice way to define an isomorphism categorically — which is somewhat obvious if you’ve seen some algebraic topology before.

Definition. A map $f:A\to B$ is an isomorphism if there is some mapping $g:B\to A$ such that $f\circ g = 1_{B}$ and $g\circ f = 1_{A}$, where $1_{A},1_{B}$ denote the identity morphism from the subscripted object to itself.

Now, naively, one might think, "Okay, if I have some certain kind of morphism in my category (set-maps, homomorphisms, homeomorphisms, poset relations, …) then if it is an epimorphism and a monomorphism, it should automatically be an isomorphism."  Unfortunately, this is not the case.  Here’s two simple examples.

Example (Mono, Epi, but not Iso).  The most simple category for which this works is the category 2, which I’ve drawn below:

There are two objects, $a,b$ and three morphisms, the identites and the morphism $f:a\to b$.  First, prove to yourself that this is actually a category.  Second, we note that $f:a\to b$ is an epimorphism: the only map from $A\to A$ is the identity, and there is no mapping from $b\to a$, so the property trivially holds.  Third, we note that $f:a\to b$ is a monomorphism for the exact same reason as before.  Last, we note that $f$ is not an isomorphism: we would need some $g: b\to a$ which satisfied the properties in the definition above…but, there is no map from $b\to a$.  Upsetting!  From this, we must conclude that $f$ cannot be an isomorphism despite being a mono- and epimorphism.

Similar Example (Mono, Epi, but not Iso).  Take the category $({\mathbb N}, \leq)$, the natural numbers with morphisms as the relation $\leq$.  Which morphisms are the monomorphisms?  Which morphisms are the epimorphisms?  Prove that the only isomorphisms are the identity morphisms.  Conclude that there are a whole bunch of morphisms which are mono- and epimorphisms but which are not isomorphisms.