# Difficulties with reading "informal" and "non-rigorous" sections when studying Algebraic topology

The words in the title may not be accurate, so I put them in quotes.

Recently I begin reading Hatcher's book on Algebraic topology. I have not studied any "topology" other than basic point set topology. I have much more experience with analysis-related topics (analysis of Banach and Hilbert spaces, for example).

I feel a lot of discomfort when reading sections of geometrical descriptions about how one space is homeomorphic to another. For example, the following paragraph taken from section 1.3 (Van Kampen's Theorem) is causing me some troubles. $$K$$ is a torus knot, embedded in $$\mathbb R^3$$. Please ignore $$X$$; it is not relevant to the discussion below. It is very easy to understand why $$\partial D^4 = \partial D^2\times D^2\cup D^2\times \partial D^2$$. This is just an exercise of point-set topology. The part after "geometrically" is, however, difficult for me. Although it is quite natural to regard $$\partial D^2\times D^2$$ as a solid torus, I cannot see any obvious reason why it can be identified with the solid torus bounded by the torus knot $$K$$. It is also very difficult to see why the complement of this solid torus is homeomorphic to $$D^2\times \partial D^2$$. (It might be not as difficult as I think, but from time to time, I get stuck on some "obvious" things.) Whenever I come across such descriptions, I struggle a lot, thinking about how to prove those statements.

Also, the sets $$D$$, $$S^3$$ and so on in the book are not specifically defined (for example, as $$D^2=\{(x,y):x^2+y^2\le 1\}$$). Although defining $$D^2$$ specifically is clearly unnecessary, sometimes, the vague concept of $$D^2$$ causes me some difficulties. I find it really hard to tell whether the author is making a rigorous statement or a non-rigorous description.

In the end, in this book, I find it much easier to read proofs than to read remarks and other descriptive sections, due to the problems outlined above. Interestingly, when I read analysis books, the opposite is true: remarks are much easier to read than proofs. I am wondering why this is the case.

How can I overcome those problems/difficulties when reading this book?

This is not seeking personal advice - I believe that this might be the feeling of many other people as well. Please ask me to clarify if anything is unclear.

• A similar question was recently asked, with no answer, but many comments were made that may be useful: Hatcher Algebraic Topology: I have all the prereqs, so why is this book unreadable for me? Jan 27, 2020 at 7:11
• this mode of writing and teaching put me off algebraic topology when I first started studying it; I was astonished at the lack of rigour. It was some years later I began to appreciate it
– T_M
Jan 27, 2020 at 12:14
• "The sets $D$, $S^3$ and so on in the book are not specifically defined..." see page xii, Standard Notations. Jan 27, 2020 at 15:02
• I find it much easier to read proofs than to read remarks and other descriptive sections, due to the problems outlined above. Well, that's just the style of the author. I would wager that in his mind, remarks are something extra that can safely be ignored, and so they require less detail. If someone is willing to learn the extra stuff, then they should be willing to put in the extra work of understanding the remark, I suppose. "Remark" is not a universally-agreed upon word: some authors take it as "please remark this, it's important!" and others as "somewhat related incidental fact". Jan 27, 2020 at 15:57
• Also, "rigorous" is not synonymous with "explicitly written down formulas". A graphical argument can be completely rigorous. Jan 27, 2020 at 16:02

Well, your example is sort of hard to read without a blackboard because a lot of different images are going on. First, it's also worth noting that the identification $$D^4=D^2\times D^2$$ is only true up to homeomorphism when working with the $$2$$-norm. Only for the $$\infty$$-norm is it an on-the-nose identity of spaces.

However, Hatcher wants to map to $$\mathbb{R}^3$$, and the canonical way of going there is via the stereographic projection from the north pole $$\phi_N((x,y,z,w))=\frac{1}{1-w}( x,y,z)$$ (when realising $$S^3$$ under the $$2$$-norm), which has inverse given by $$(x,y,z)\mapsto \frac{2}{x^2+y^2+z^2+1}(x,y,z,0)+\frac{x^2+y^2+z^2-1}{x^2+y^2+z^2+1}(0,0,0,1)$$. Note that $$\phi_N^{-1}$$ extends continuously to the $$1$$-point compactification by mapping $$\infty$$ to $$(0,0,0,1)$$.

Okay, so what's $$S^1\times S^1$$ in $$S^3$$ under the $$2$$-norm? Well, it's any set of the form $$\{(x,y)\in(\mathbb{R}^2)^2| \|x\|_2=r,\|y\|_2=s\}$$ with $$r^2+s^2=1$$. For simplicity, set $$r=s=\frac{1}{\sqrt{2}}$$. For such an element $$(x_1,x_2,y_1,y_2),$$ we see that, for fixed $$y_2$$, the image of $$\phi(x,y)$$ is two circles for $$|y_2|\neq \frac{1}{\sqrt{2}}$$ and a unique circle for $$|y_2|=\frac{1}{\sqrt{2}}$$ (since this forces $$y_1=0$$). Now, at the same time, we see that, as $$y_2$$ ranges over $$[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}],$$ the third component of the image ranges over an interval (by continuity). Hence, $$\phi(S^1\times S^1)$$ is, indeed, the usual torus shape. Hence, $$\mathbb{R}^3\setminus \phi_N(S^1\times S^1)$$ has two components. Let $$U$$ denote the "inside", i.e. the bounded component, the closure of which is homeomorphic to the solid torus $$\partial D^2\times D^2$$. Hence, so is $$\phi^{-1}(\overline{U})$$. However, consider the stereographic projection $$\phi_S$$ from the south pole instead, and you'll see that it completely flips the picture, i.e. $$\phi_S(\phi_N^{-1}((\mathbb{R}^3\cup \{\infty\})\setminus U)=\overline{U}$$

Hence, the outside of the torus (with $$\infty$$), is homeomorphic to the inside of the torus, which exactly gives you Hatcher's claim.

As for generally understanding intuitive explanations, I really recommend trying to draw situations. Naturally, drawing in four dimensions is pretty hard, but you can actually make a drawing in $$3D$$, where you end up cutting the top and bottom of $$S^2$$ (which gives you a copy of $$\{0,1\}\times D^2$$), which leaves you with a middle part (homeomorphic to $$[0,1]\times S^1$$).

As a finishing anecdote, a wise, older student once told me: "There are two kinds of people who do algebraic topology: The ones that love all the diagrams, and try to ignore the geometric pictures, and the ones that love weird geometry and decide to live with all the diagrams." I think that sort of explains the weird interplay between the abstraction and the handwaving.

• Nice answer and nice username (+1). Mar 6, 2021 at 21:24

The answer of @WoolierThenThou is good for motivation and intuition. I want to explain something a bit different, namely how to read advanced mathematics books, and the hard work that is sometimes (often?) needed.

These books have heavy prerequisites, and you have to know those prerequisites. In the case of Hatcher's textbook, there are very heavy prerequisites in analytic geometry and topology. If you are not good in analytic geometry and/or topology, then you have to be ready to back up and learn what you need to know. What you wrote about your background gives me a clue: "I have not studied any 'topology' other than basic point set topology"; well, then you're going to need to study some topology. Picking up Hatcher knowing only some point set topology is something like picking up a book on Banach and Hilbert spaces knowing only some advanced calculus.

For example, let's take a phrase which comes up on the page preceding the one you are reading:

... the torus $$S^1 \times S^1$$ is embedded in $$\mathbb R^3$$ in the standard way.

The author is definitely assuming a strong prerequisite here: not just that you are familiar with $$S^1 \times S^1$$ and $$\mathbb R^3$$ as topological spaces, but that you also know standard embedding of $$S^1 \times S^1$$ in $$\mathbb R^3$$.

If you don't know that embedding, then yes, you're gonna be stuck, confused, etc. So the correct thing for you to do is put down that book and go figure out what embedding the author is referring to. It's an example in topology, so more elementary books on analytic geometry or topology might be a good place to start.

For instance, you might be able to recall an example from analytic geometry or multivariable calculus: a torus is often depicted as the surface of revolution obtained by revolving a certain circle in the $$x,z$$ plane around the $$z$$-axis in $$x,y,z$$ space, for instance the circle $$(x-2)^2 + z^2 = 1$$. And now you might be able to use that description, combined with your expertise in analytic geometry, to either write down a parametric formula $$x=f(s,t)$$, $$y=g(s,t)$$, $$z=h(s,t)$$ for the torus, or perhaps to write down an equation $$F(x,y,z)=0$$ whose solution set is that that torus (cylindrical coordinates help). And now you might be able to use, or develop, your expertise in elementary topology to convince yourself that this torus embedded in $$\mathbb R^3$$ that you've just specified is indeed homeomorphic to the product space $$S^1 \times S^1$$.

Now that you've got that settled, read on. Let's turn to something else from your question. You wrote in your post:

Although it is quite natural to regard $$\partial D^2\times D^2$$ as a solid torus, I cannot see any obvious reason why it can be identified with the solid torus bounded by the torus knot $$K$$.

Here a big part of the trouble looks like you simply misread the passage from Hatcher's book, which I'll quote:

The first solid torus $$S^1 \times D^2$$ can be identified with the compact region in $$\mathbb R^3$$ bounded by the standard torus $$S^1 \times S^1$$ containing $$K$$.

Notice, we are not talking about the "solid torus bounded by the torus knot $$K$$", instead we are talking about "the ...solid torus... in $$\mathbb R^3$$ ... bounded by the standard torus $$S^1 \times S^1$$..."

Well, assuming that you have just carefully verified the standard embedding of $$S^1 \times S^1$$ into $$\mathbb R^3$$ as just described, you should now have little trouble extending this to an embedding of the solid torus into $$\mathbb R^3$$: instead of simply revolving the circle $$(x-2)^2 + z^2 = 1$$, revolve the entire disc $$(x-2)^2 + z^2 \le 1$$.

There are plenty of other things you will encounter, one of which is nicely explained in the answer of @WoolierThanThou, namely the standard homeomorphism between $$S^3$$ and the one-point compactification of $$\mathbb R^3$$; if you don't know that homeomorphism, stop, put down Hatcher's book, and learn about it.

Once you know that homeomorphism, then you can continue on to "the second solid torus..."

• The thing is that although I know every thing you mentioned, I find it quite time consuming and difficult to go over all the details. And non of differential/analytic geometry is explicitly claimed to be a prereq either in the book or in the description of a course based on this book. Jan 27, 2020 at 23:05
• In analysis books, it seems that results from prereqs are usually explicitly stated. Jan 27, 2020 at 23:07
• @Jethro Analytic geometry is not the same as differential geometry. Jan 27, 2020 at 23:26
• @Jethro Filling out details is time consuming and difficult. In general, I find it much easier to just internalise ideas and general intuition. For instance, it took quite a bit of time to work out my above answer (partly because, at first, I read the Hatcher text wrong). Why? Well, even if, in my brain, I had the mindmap "use stereographic projection to map the torus to the torus, flip the stereographic projections to identify inside and outside of torus," I still had to recall how the hell the stereographic projection looks in formula. Jan 28, 2020 at 8:29
• And more often than not, as you start filling out details, it often happens that you realise that there is some detail you've glossed over - at least this happens to me all the time. If you have good people to talk to regarding the general concepts, then the internalisation just becomes much easier, because the people in question can help point out what the important moral lessons are. In general, this is very hard to get from a textbook, and when you do find one that can - Rudin II springs to mind for myself - it's quite extraordinary. Jan 28, 2020 at 8:32