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What kind of topological questions does algebraic topology answer where point set topology is not enough?

Phrased differently:

Where is the line (or maybe intersection) between point set topology and algebraic topology? What is the distinction between the problems each deals with?

I want to understand the motivation behind introducing the algebraic machinery to topology.

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    $\begingroup$ At the simplest level, it provides a method to show two topological spaces are not homeomorphic. For example, how would you prove ${\mathbf R}^m$ is not homeomorphic to ${\mathbf R}^n$ when $m \not= n$? You can distiguish ${\mathbf R}$ from ${\mathbf R}^n$ for $n > 1$ since removing any point from ${\mathbf R}$ makes it disconnected, but that's not so for ${\mathbf R}^n$ when $n > 1$. However, to handle the case $m > 1$ and $n > 1$ is not as easy. But with homology groups you can distinguish these spaces since the sequence of homology groups of ${\mathbf R}^n$ determines $n$. $\endgroup$ – KCd Jan 9 '14 at 23:01
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    $\begingroup$ @user116457: Both general and algebraic topology are ultimately interested in determining which topological spaces are homeomorphic. The algebraic topologist generally achieves this by translating topological questions into algebraic questions, while hoping that the algebraic questions are easier to answer. While not an expert, my belief/understanding is that the classes of spaces that allow for the common translations into algebraic terms are fairly restricted (at least compared to the class of all topological space). $\endgroup$ – user642796 Jan 9 '14 at 23:11
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    $\begingroup$ It's REALLY PRETTY. Who would ever have guessed, pre-1900, that associated to a topological space is a whole suite of algebraic invariants? It's a deep, rich connection between two fields which seem, at first glance, to be entirely unrelated and opposite each other. $\endgroup$ – Neal Jan 9 '14 at 23:11
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    $\begingroup$ Historically, Algebraic Topology has been used to prove some geometrically interesting theorems. For example, one of the first applications of K-theory was to give a nice proof of when $\mathbb{R}^n$ is a division algebra, or equivalently, then the tangent bundle to $S^n$ is trivial. Another big one is the Index Theorem. I think the motivation originally came from the fact that it was able to solve some outstanding problems in mathematics and so was itself quite interesting. $\endgroup$ – Moss Jan 9 '14 at 23:12
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    $\begingroup$ @KCd: good points but your comment is quite imprecise. All ${\bf R}^n$ have homotopy type of a point, so that their homology and actually all algebraic topological properties coincide. What's different is e.g. the homology of ${\bf R}^n \setminus {\rm pt}$ which has the homotopy type of a sphere. Another way would be using compactly supported cohomology but these invariants are of course not preserved by homotopy. $\endgroup$ – Marek Jan 10 '14 at 14:21
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The best elementary justification for algebraic topology is that it provides a way of rigorously proving when some spaces are not homeomorphic. It's intuitively obvious, for instance, that the 2-dimensional sphere $S^2$ and the torus $T^2$ are very much different spaces. But to actually prove from scratch that every map $S^2\to T^2$ is either not bijective or has a discontinuous inverse is daunting. On the other hand, as soon as you know the first facts of homology theory or about the fundamental group you can write down a proof in just a couple of lines.

A somewhat more sophisticated motivation is that algebraic topology has the best tools for making sense of questions that are invariant up to homotopy-most simply, which spaces are homotopy equivalent. The fundamental group (and higher homotopy groups) and (co)homology theory (theories) are just as good at distinguishing non-homotopy equivalent spaces as non-homeomorphic ones, which at first sight is even harder to do. Now, you'll notice I haven't said anything about proving spaces are homeomorphic, or homotopy equivalent. The former is impossible in generality much beyond, say, 2-dimensional manifolds. But the latter is possible, at least in theory, for a kind of space called CW-complexes, which are thus a favorite of algebraic topologists.

The point, overall, is that algebraic topology provides one with discrete invariants that are more tangible material for writing rigorous proofs than the purely topological motivation for an idea-it's much harder to fully comprehend a space, especially in words, than a group associated to that space. This reflects the common pattern that algebra is more verbal and geometry more visual.

I can't comment very well on the dividing line between general topology and algebraic topology, because I don't know anything at all about modern topological research that's not either algebraic or geometric (i.e. about manifolds.)

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  • $\begingroup$ Very good answer, I agree with most of what you have said. But your example $S^2 \to T^2$ is actually pretty simple. Just write $S^2$ as union of hemispheres. Image of the equator must be contractible, so that image of one of the hemispheres is then a disk, while the image of the other one contains two great circles (generators of homology of $T^2$) and wouldn't be contractible. $\endgroup$ – Marek Jan 10 '14 at 9:50
  • $\begingroup$ Even more than that, maps between Riemann surfaces (and even branched covers) were understood almost a century before algebraic topology appeared. I think the invariance of domain might be a better example (although I'm not sure it doesn't have a simple proof without using AT). $\endgroup$ – Marek Jan 10 '14 at 10:02
  • $\begingroup$ Thanks for the comments, Marek. Your proof does use the fact that the generators of the torus' homology aren't contractible, though, which I'm thinking of as essentially an algebraic topology fact since I would prove it via Mayer-Vietoris. I didn't realize that Riemann surface theory was developed so early as that, though, thanks for the information. $\endgroup$ – Kevin Carlson Jan 10 '14 at 13:24
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In order to "understand the motivation behind introducing the algebraic machinery to topology" you need to go back to the history of the subject, and how it developed out of problems in complex analysis, as did general topology too. If you can get hold of it, I recommend "History of Topology", edited I M James, Noth Holland, 1999, particularly as a start the articles by S. Lefschetz on "The early development of algebraic topology" and by I M James on "From combinatorial topology to algebraic topology". I also recommend the Introduction to Lefschetz's "Introduction to topology" (but not the rest of the book, which many find confusing). You should not think that general topology arose first, but instead that they developed together.

A problem with the early work was to define precisely "boundaries" and "cycles", to obtain the rule "every boundary is a cycle"; this is the rule $\partial \partial =0$. It was Poincar\'e who developed the excellent trick of taking formal sums of oriented simplices in a simplicial complex. It was much later that Eilenberg introduced the ordered simplex and the rule we know and love $$ \partial = \sum_{i=1}^n (-1)^i \partial _i .$$ In fact the idea of taking "formal sums" derives from integration over various domains, i.e. of writing for convenience $$\int_{C} f + \int_D f = \int_{C +D} f .$$ The story is complicated and with many twists, and you should not think the story of conceptual development has ended. I do agree that the question is a good one, since the intuitions of earlier generations may not have been fully expressed in the current formulations. For example, the vision of the topologists of the beginning of the 19th century idea of higher dimensional nonabelian versions of the fundamental group was dismissed from the 1930s and on, essentially because of the so-called Eckmann-Hilton argument, showing "double groups" are just abelian groups. However it turns out that "double groupoids" are more complicated than groups, and this idea has quite a long way to run!

I mention that the word "groupoid" does not occur in James' book.

January 14, 2017 There is more discussion in this preprint.

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