The fundamental group functor part 2

So... in the previous post I promised to finish what I was saying about the fundamental group functor... So far I've sketched the proof that this is, indeed, a functor. I would show the proof in detail, but it's long, tedious and not very informative-- the point is, it's a map from Top$_*$ to Grp which preserves morphisms. There are much more interesting things a functor can preserve. Namely, it can preserve products and coproducts.

So what's a product? As a motivating example, look at Set. When we talk about the product of two sets, we clearly mean the cartesian product. Since we're interested in category theory at the moment, we don't really want to talk about members of the product, we want to talk about maps to and from the product.

It turns out there are two really nice maps $pr_1:A\times B\rightarrow A$ and $pr_2: A\times B\rightarrow B$, the projections onto $A$ and $B$ respectively. It turns out that they have a really nice universal property: Given an object $V$, and two maps $f:V\rightarrow A$, $g:V\rightarrow B$, we can "factor" $f$ and $g$ through $A\times B$ in a unique way. This means that we have a unique map $h:V\rightarrow A\times B$ such that $f=pr_1\circ h$ and $g=pr_2\circ h$. At first it may be a bit surprising that this map (sometimes called $f\times g$) is unique. But really, our two projections forget everything about one side of our product, so the function needs to act "independently" on $A$ and $B$, and there's really only one way to get this to interact properly with the projections.

Something that's more surprising is this: the product is unique up to unique isomorphism. This means that if there is a "different" product (Why not try $B\times A$?), there is a single, canonical isomorphism between the two objects-- just factor the projections from one product through the other. This map is unique, and it damned-well better be an isomorphism. (To see that it is, factor the projections back the other way, wave your hands about and say something about "the identity morphism".)

Ok. So, by analogy with Set, we (sort of) get what a product is. What about coproducts? A nice thing about category theory is that whenever you see a word that starts with "co", you can figure out what it means in 3 easy steps:

1. remove the "co" from the word.
2. Draw the diagram that represents the word you just found.
3. Turn around all the arrows.

So, this means the coproduct, $A\coprod B$, should have two maps $i_1:A\rightarrow A\coprod B$ and $i_2:B\rightarrow A\coprod B$ (called the imbeddings) such that for any pair of maps $f:A\rightarrow V$ and $g:B\rightarrow V$, we have a unique morphism $h:A\coprod B\rightarrow V$ such that $h\circ i_1 = f$ and $h\circ i_2 = g$. For some reason this always seems a little harder to follow. Let's work it out in Set. Let's look at $f$ and $g$ as in the definition. We want some map (call it $f*g$ because I can't think of what the actual notation is) that goes from somewhere to $V$ such that $f*g \circ i_1 = f$ and the same with $g$. We want $i_1$ and $i_2$ to do almost nothing... what happens if we take $A\coprod B$ as the disjoint union? (Hence the notation...) What is the imbedding? It's the "move me from $A$ to $A\coprod B$" function. And what could $f*g$ possibly be? Well obviously, it's the function which sends $a\in A\mapsto f(a)$ and $b\in B\mapsto g(b)$.

Ok. Great. We know what the product and coproduct need to look like (at least when we only care about the product of two objects). What exactly are they in Top$_*$? It turns out they are the wedge products-- take the disjoint union (familiar?) and glue the two spaces together at their base-points. This means that we have two completely unrelated spaces (modulo open sets containing the basepoint.)

This idea of a coproduct being the result of "smashing together" two objects without making them at all related is basically consistent throughout basically every category. In fact, in the category of groups, it's the free product, which is the "freely generated" product of the two groups-- For groups $G$ and $H$, this means the set of all words on $G\cup H$, where things reduce in the "obvious" way and no other way... I'm going to pretend like this makes sense to you, since (as you've surely learned by now) I have yet to decide what level of audience I'm writing for.

So, taking this back to the fundamental group functor: for two (pointed) spaces $(S,s)$ and $(T,t)$, we would like $\pi_1(S\times T, (s,t)) = \pi_1(S,s)\times \pi_1(T,t)$ and $\pi_1(S\coprod T, (s,t)) = \pi_1(S,s)\coprod \pi_1(T,t)$.

Guess what? I'm going to cop out of actually proving this! (Are you surprised? You should be used to this by now...)

However, I will at least wave my hands around a bit and give you a feel for why it's true. First let's look at products. As an example, look at the torus-- $S^1\times S^1$. Draw a path on this. We want to be able to push this path down to a path which only lives in one copy of $S^1$ in each component. (I.e., a path which stays on $(S^1\times\{0\})\cup(\{0\}\times S^1)$. ) We can do this by pushing (in a continuous fashion-- i.e., homotopically) all points of our path onto one of our two reference circles.

For coproducts: it's a little more obvious in some sense--- any path is going to stay in one of our two spaces for a while, and then cross over to the other. The homotopy group we get here "reduces" in the obvious way, and no way else-- i.e. it's the free product.

Ok. there. I've fulfilled my promise. Expect a more detailed and less obnoxiously hand-wavy post about natural transformations and path categories soon (TM)

The Zahir and Asterion

Sorry... I know I promised to finish that last post about 2 weeks ago... I'll get around to that soon. In the mean time, I've just started reading The Zahir by Borges, which I somehow haven't read. I could have sworn I had read the whole of The Aleph, but I digress. I stumbled upon the following passage (I don't know who the translator is):

Until the end of June I distracted myself by composing a tale of fantasy. The tale contains two or three enigmatic circumlocutions: “water of the sword”, it says, instead of blood, and “bed of the serpent”, for gold, and is written in the first person. The narrator is an ascetic who has renounced all commerce with mankind and lives on a moor. (The name of the place is Gnitaheidr.) Because of the simplicity and innocence of his life, he is judged by some to be an angel; that is a charitable sort of exaggeration, because no one is free of sin. He himself (to take the example nearest at hand) has cut his father’s throat, though it is true that his father was a famous wizard who had used his magic to usurp an infinite treasure for himself.

Protecting this treasure from mad human greed is the mission to which the he has devoted his life; day and night he stands guard over it. Soon, perhaps too soon, that watchfulness will come to an end: the stars have told him that the sword that will cut him off forever has already been forged. (Gram is the name of the sword.) In an increasingly tortured style, the narrator praises the luster and flexibility of his body; one paragraph offhandedly mentions “scales”; another says that the treasure he watches over is of red rings and gleaming gold. At the end, we realize that the ascetic is the serpent Fafnir and the treasure on which the creature lies coiled is the gold of the Nibelungen. The appearance of Sigurd abruptly ends the story.

This sounds rather amusingly like... The House of Asterion which was published in the same collection. This is one of the things I really like about Borges: He makes very subtle references to other works of his. Can anyone think of any other specific examples of this?

The fundamental group functor

This is something I've always (read: since I learned about it less than 6 months ago) found pretty neat. There's nothing terribly original here-- everything can be found in any algebraic topology book, and in most general topology books, but I don't think categorical language makes its way in there all the time...

The point of this "little" post is to point out that the operation taking a (pointed) topological space $(X,x_0)$ to it's fundamental group, $\pi_1(X,x_0)$ is a functor which preserves products and coproducts... (Did that sentence have a point? Sorry... I'm done.)

First, as a technical point: we need to work int he category of pointed spaces: Top$_*$. (A pointed topological space is just a pair $(X,x_0)$ where $x_0\in X$. The morphisms are continuous functions $f:(X,x_0)\rightarrow (Y,y_0)$ such that $f(x_0)=y_0$. The idea is we are distinguishing a point, just as we do to get the fundamental group.) The reason for this is that it gives a nice way of distinguishing between base points (for our fundamental group) in different path-components-- every selection of base point gives us a new space-- Some are isomorphic. This allows us the avoid the technical nightmare of what to do with non-path-connected spaces. (I.e., we don't get a functor if we're only working in Top) There's another reason for this: Wedge products give Top$_*$ a sensible notion of coproduct-- or at least, one which is actually preserved by the functor.

So, first of all, what does it mean for us to have a functor? A functor is a map between categores which preserves identities and composition of morphisms. In other words, for categories $C$ and $D$, $F:C\rightarrow D$ is a functor if $F(id_c)=id_{F(c)}$ for every object $c\in C$, and for every pair of morphisms
\[c_0\stackrel{f}{\rightarrow}c_1\stackrel{g}{\rightarrow}c_2\]
In C, we have that $F(g)\circ F(f) = F(g\circ f)$.

Given a function $f:(X,x_0)\rightarrow(Y,y_0)$, $f$ induces a homomorphism $f_* : \pi_1(X,x_0)\rightarrow \pi_1(Y,y_0)$-- Any path in $X$, when fed through $f$ becomes a path in $Y$. Since the map preserves basepoints, a loop at $x_0$ becomes a loop at $y_0$-- seeing that this is compatible with homotopy isn't too difficult.

To say that $\pi_1(-)$ is a functor means that $f_*(\pi_1(X,x_0)) = \pi_1(\operatorname{Im} f,y_0)$ and that $(id_X)_* = id_{\pi_1(X)}$ (Sorry, commutative diagrams are not working so hot in this $\LaTeX$ package... I'll need to do something about that.) A quick diagram chase shows that this is the case.

Now is where things finally get interesting... and... I'm tired, and will finish this later today.

A few words on Balaam's Error

I'm not sure why I'm writing this down now, but: I don't agree that "Balaam's error" has anything to do with money. Based on his actions, monetary reward seems to be a small concern for him. His error comes from this: He is afraid to contradict the Moabites. He is too polite, too unwilling to offend.
Sometimes things need to be said which are offensive-- causing offense is rarely good in its own right, but offensive things are important. Balaam was too afraid (either socially, or for his life) to tell the Moabites something offense, "God says 'no!'"

We should learn from this.

Balaam

I wrote this sometime last year, but apparently forgot about it.
***

I tread on with the Moabites, seeking their praise, silver and jewels. "I cannot contradict the ineffable," but I march on, afraid to contradict these messengers.

But still my ass, wiser than I, stubbornly refuses to move. Twice she's flogged, and twice she stands up and walks. Again she's flogged, but finally she speaks her mind:
"Why do you treat me so? Have I not carried you from your home?"
And thus I lie: "No."

Nuclear energy

is safe and clean. Ask anyone who knows anything about it.

Mathematical insight...

This is from a reply I posted here to a question about gaining mathematical insight.

* Nothing is "obvious".
Try to be extremely formal with all of your proofs. Make sure your steps all follow immediately from previous steps, definitions or theorems. Spend some time proving the "really basic" properties that follow immediately from applying the definition. Also, ask yourself what sort of objects satisfy certain properties, and which don't. Eg. For complete metric spaces, come up with a "canonical" example of a complete metric space, a "canonical" incomplete metric space, and a degenerate example of each. For example, the discrete metric is complete (if you know about metric spaces, you may want to prove this), but it really doesn't match our intuition for what a complete metric space "should be."

On that note, try to understand what the intuition for a property or object is-- what does it "mean" for a set to be a group under an operation? Also, try to keep track of where intuition departs from math-- For example, we like to think of topological spaces geometrically, but there are some very non-geometric topological spaces.

* Rewrite the same thing as many different ways as you can.
For example, if the problem asks a question about a normal subgroup, you should be thinking of all the characterizations of normality-- It's the kernel of a homomorphism, it's invariant under conjugatian (which really is the same as its left and right cosets are the same), if a and b are in the same coset of N, then a-b is in N.

* When working on a proof, pay attention to everywhere you use your assumptions.

* After writing a proof, make sure the result seems to make sense.
Does it match up with intuition? If not, figure out why. If the problem is with your intuition, try to figure out what you are assuming to be true, and make a note of it.

Are any basic examples of the structure a counter-example to your "theorem"? Does each step follow from the last? Are you sure?
(I have a friend who has written 3 or 4 wrong proofs this semester, and every time, he realized it was wrong based on these checks, although normally I had to pick out the false step for him :D )

* Learn to look for counter-examples.
If you're asked to prove something wrong, look at some basic examples of the structure you're looking at. Does the statement hold for them? If so, can you see what properties make it work? If so, try to come up with an example where that property doesn't hold. Does the statement fail now? Rinse and repeat.

* Rewrite your assumptions. Rewrite them in different words. Rewrite them with the definitions of any terms you are uncomfortable with.

* Look for connections.

* Rewrite any objects you're looking at in terms of other objects. The complement of an open set is closed. The complement of a closed set is open. A connected space has proper (non-empty) clopen sets. g is in the Center of G means gh=hg for any h.

* State the obvious. Often. And then state it again.

* Ask stupid questions. Then answer them.
Is R complete? Why is a polynomial continuous? Is Z abelian? Finitely generated? What about Z^n? What does Abelian mean anyway?

* Don't be afraid to ask someone else stupid questions.

* Don't be discouraged when you sit for hours without understanding what to do; let the gears keep grinding.
Put on some music and rock out while you think. Rewrite the assumptions. Try to do something. When you get stuck, try to figure out why that doesn't work. Does it get you anywhere at all?

* Don't be afraid to go do something else for an hour or 2 and then come back to work on a problem.
This is when some of the best insights happen-- go make some tea, read a book, watch a movie, get coffee with a friend, do something. Then come back and start again. Sometimes it'll be hard to get back in the zone-- redo some easier problems: Try to reword your argument or try to find a cleaner argument.

* Work on a simpler problem.
Need to separate two compact sets? Don't! separate a compact set from a point. Can you use this same argument again? Will a similar argument work for two sets?

* Work on a more general problem.
Don't show that n is divisible by 3, show that all numbers of a certain form are divisible by 3. Then show that n has that form.

Hope these give you something useful to think about.