4 ms·

Did anyone suggest thinking of ker/im as "boundaries modulo cycles"? That is, homology detects "holes," loosely speaking, and a hole is exactly something that y

Did anyone suggest thinking of ker/im as "boundaries modulo cycles"? That is, homology detects "holes," loosely speaking, and a hole is exactly something that you can draw a cycle around, where that cycle is not the boundary of some disk (in 2 dimensions, for example). To be even more concrete, if I cut out a disk from a piece of paper, I've created a cycle (the circle bounding the disk) that is not the boundary of any disk in the piece of paper (because I cut it out). Admittedly, this gets a little funkier in higher dimensions, but for 2 dimensions, it's a good guide. (I think Hatcher's book has a discussion of how to make this idea precise in higher dimensions.)

prionassembly 6y ago
It was the "modulo" I had trouble with (the idea of quotient spaces). I understood well how kernels corresponded to cycles (people on IRC made me do lots of calculations with chain complexes both in Z/2 and R), and had some idea of groups (or, at least, of when something is a semigroup and not a group...) but didn't have an internal feeling of what quotient groups were.
- jvvw 6y ago
  Do you grok equivalence classes because quotients (of anything) are basically sets of equivalence classes? And on the parenting front, luckily you start with a baby not a ten year old so you have lots of chance to grow as a parent before making those sorts of decisions - a lot of parenting is deciding when to be directive/non-directive and when to take charge based on your superior understanding of the world (not letting your toddler walk into the road!) and when to let them learn from their own mistakes.
Y_Y 6y ago
This is just the same as my own conception, then again I also used (and can highly recommend) Hatcher. He kindly provides a pdf on his webpage here: https://pi.math.cornell.edu/~hatcher/AT/ATpage.html https://pi.math.cornell.edu/~hatcher/AT/ATpage.html
bikenaga 6y ago
Small correction - it's "cycles mod boundaries". In general, a quotient object R/I is read "R mod I" (top mod bottom). What's a cycle? It's something in the kernel of the boundary map. To be in the kernel means the boundary map takes it to 0. So heuristically, cycles are "things with no (zero) boundary". For the quotient cycles/boundaries to make sense - for any quotient to make sense - the bottom should be contained in the top. Like in Z/6Z, the integers mod 6, the "6Z" (the integers which are a multiple of 6) is contained in Z (the integers). In this case, for "cycles/boundaries" to make sense, a boundary should be a cycle - that is, a boundary should have no boundary. This is (again heuristically) what the definition of a boundary map (d^2 = 0) means. Consider your disk/circle analogy. The circle is a boundary (the boundary of the disk). Does the circle have a "boundary"? No - the boundary of a boundary is 0. When the quotient cycles/boundaries is nonzero, you have nontrivial homology. Think of nontrivial homology as "having a hole". To see an example of this, take the plane but remove the origin (0, 0). Consider your circle again (say the unit circle x^2 + y^2 = 1). It's still a cycle, because its boundary is 0 (no boundary). But it's not the boundary of "something" (where "something" means roughly "something nice" - it's hard to explain this better without getting technical). So the cycle represented by the circle represents a nontrivial homology class (in the homology of plane minus origin).

prionassembly 6y ago

It was the "modulo" I had trouble with (the idea of quotient spaces). I understood well how kernels corresponded to cycles (people on IRC made me do lots of calculations with chain complexes both in Z/2 and R), and had some idea of groups (or, at least, of when something is a semigroup and not a group...) but didn't have an internal feeling of what quotient groups were.

jvvw 6y ago
Do you grok equivalence classes because quotients (of anything) are basically sets of equivalence classes? And on the parenting front, luckily you start with a baby not a ten year old so you have lots of chance to grow as a parent before making those sorts of decisions - a lot of parenting is deciding when to be directive/non-directive and when to take charge based on your superior understanding of the world (not letting your toddler walk into the road!) and when to let them learn from their own mistakes.

Y_Y 6y ago

This is just the same as my own conception, then again I also used (and can highly recommend) Hatcher. He kindly provides a pdf on his webpage here: https://pi.math.cornell.edu/~hatcher/AT/ATpage.html https://pi.math.cornell.edu/~hatcher/AT/ATpage.html

bikenaga 6y ago

Small correction - it's "cycles mod boundaries". In general, a quotient object R/I is read "R mod I" (top mod bottom). What's a cycle? It's something in the kernel of the boundary map. To be in the kernel means the boundary map takes it to 0. So heuristically, cycles are "things with no (zero) boundary". For the quotient cycles/boundaries to make sense - for any quotient to make sense - the bottom should be contained in the top. Like in Z/6Z, the integers mod 6, the "6Z" (the integers which are a multiple of 6) is contained in Z (the integers). In this case, for "cycles/boundaries" to make sense, a boundary should be a cycle - that is, a boundary should have no boundary. This is (again heuristically) what the definition of a boundary map (d^2 = 0) means. Consider your disk/circle analogy. The circle is a boundary (the boundary of the disk). Does the circle have a "boundary"? No - the boundary of a boundary is 0. When the quotient cycles/boundaries is nonzero, you have nontrivial homology. Think of nontrivial homology as "having a hole". To see an example of this, take the plane but remove the origin (0, 0). Consider your circle again (say the unit circle x^2 + y^2 = 1). It's still a cycle, because its boundary is 0 (no boundary). But it's not the boundary of "something" (where "something" means roughly "something nice" - it's hard to explain this better without getting technical). So the cycle represented by the circle represents a nontrivial homology class (in the homology of plane minus origin).