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Jan 27, 2021 at 19:18 comment added Noah Snyder Slide 4 of this talk of Stillwell's suggests that the original definition of Betti numbers was the number of "cuts" by curves, surfaces, etc. needed to reduce the manifold to a simply connected region.
May 22, 2017 at 13:42 comment added John Baez I'm not sure how Betti defined his numbers, but the dimensions of images and kernels can be studied without thinking of them as dimensions of vector spaces, so maybe that's what he did. For example, the "dimension of the image of a linear transformation" can also be thought of as "how many nonzero rows your matrix has after you do Gaussian elimination", and the latter concept probably came first historically.
Dec 25, 2009 at 11:59 vote accept Gil Kalai
Dec 25, 2009 at 11:59 history bounty ended Gil Kalai
Nov 11, 2009 at 17:54 comment added Gil Kalai I came across this description, but I cannot imagine how Betti numbers were defined to start with.
Nov 11, 2009 at 17:50 comment added Scott Carter Yes, that has been explicitly discussed in several of Oleg Viro's talks about Khovanov homology.
Nov 11, 2009 at 17:42 comment added Reid Barton Today they are, but apparently hthere must have been another definition in the 1920s, since the insight of Emmy Noether that the Betti numbers are the ranks of certain groups was a big deal back then: en.wikipedia.org/wiki/Emmy_noether#Contributions_to_topology
Nov 11, 2009 at 17:10 comment added Gil Kalai what do you mean by moving from Betti numbers to homology; arnt the Betti number defined as the dimensions of homology to start with?
Nov 11, 2009 at 16:51 comment added Reid Barton Yes, or maybe even better would be to say moving from Betti numbers to homology.
Nov 11, 2009 at 10:34 comment added Gil Kalai Is moving from the notion of "Euler characteristic" to the notion of "homology" a sort of prototype to categorification?
Nov 11, 2009 at 2:14 history edited Scott Carter CC BY-SA 2.5
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Nov 11, 2009 at 0:28 comment added alekzander second paragraph: subject to the relation $d\circ x=x\circ d+1$, of course
Nov 10, 2009 at 14:54 history answered Scott Carter CC BY-SA 2.5