## 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.

## Orthogonal Complement of Even Functions.

### August 5, 2012

Question:  Consider the subspace $E$ of $L^{2}([-1,1])$ consisting of even functions (that is, functions with $f(x) = f(-x)$).  Find the orthogonal complement of $E$.

One Solution.  It’s easy to prove $E$ is a subspace.  Then, there is a representation of any function in this space by adding odd and even functions together; more precisely, given $f\in L^{2}([-1,1])$ we have that $\frac{f(x) + f(-x)}{2}$ is even and $\frac{f(x) - f(-x)}{2}$ is odd and $f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$.  For uniqueness, note that if $f(x) = f(-x) = -f(x)$, then $f(x) = -f(x)$ for each $x$, giving us that $f(x) = 0$.  Hence, the orthogonal complement of $E$ is the set of odd functions.  $\diamond$

Here’s another solution that "gets your hands dirty" by manipulating the integral.

Another Solution.  We want to find all $g\in L^{2}([-1,1])$ such that $\langle f, g \rangle_{2} = 0$ for every even function $f\in L^{2}([0,1])$.  This is equivalent to wanting to find all such $g$ with $\displaystyle \int_{-1}^{1}f(x)g(x)\, dx = 0$.  Assume $g$ is in the orthogonal complement.  That is,

$\displaystyle 0 = \int_{-1}^{1}f(x)g(x)\, dx = \int_{-1}^{0}f(x)g(x)\, dx + \int_{0}^{1}f(x)g(x)\, dx$

$\displaystyle = -\int_{1}^{0}f(-x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx$

The last equality here re-parameterizes the first integral by letting $x\mapsto -x$, but note that our new $dx$ gives us the negative sign.

$\displaystyle = \int_{0}^{1}f(-x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx$

$\displaystyle = \int_{0}^{1}f(x)g(-x)\, dx + \int_{0}^{1}f(x)g(x)\, dx$

$\displaystyle = \int_{0}^{1}f(x)g(-x) + f(x)g(x)\, dx = \int_{0}^{1}f(x)(g(x) + g(-x))\, dx$.

We may choose $f(x) = g(x) + g(-x)$ since this is an even function, and we note that this gives us

$\displaystyle 0 = \int_{0}^{1}(g(x) + g(-x))^{2}\, dx$.

Since $(g(x) + g(-x))^{2}\geq 0$, it must be the case that $(g(x) + g(-x))^{2} = 0$.  [Note: The fact that this is only true "almost everywhere" is implicit in the definition of $L^{2}$.]  Hence, $g(x) + g(-x) = 0$, giving us that $-g(x) = g(-x)$

We now have one direction: that if $g$ is in the orthogonal complement, then it will be odd.  Now we need to show that if $g$ is any odd function, it is in the orthogonal complement.  To this end, suppose $g$ is an odd function.  Then by the above, we have

$\displaystyle \langle f, g \rangle_{2} = \int_{-1}^{1}f(x)g(x)\, dx = \int_{0}^{1}f(x)(g(x) + g(-x))\, dx = 0$

where the last equality comes from the fact that $g$ is odd.  $\diamond$

## Soft Post: Tools of the Trade.

### January 26, 2011

Before this post, I outlined another post which attempted to find the average price of being a mathematician — or, in other words, what does it cost to be able to get a BA in math, including tuition, tools, books, and so forth.  A lot of my estimates were arbitrary, so I’ll post about a slightly different topic: what tools do I use on a regular basis to do mathematics, and how much do they cost?