@BalarkaSen So, at the end of the previous chapter, there's an exercise to show that the Euler characteristic (defined in terms of Betti numbers) is equal to the CW-complex definition
That is, where $\chi(X):=\sum_i(-1)^ib_i(X)$, prove that $\chi(X)=\sum_i(-1)^i\alpha_i$ where there are $\alpha_i$ cells in dimension $i$
Any hints on how to start that? (This was at the end of the cellular homology chapter.)
Also, @BalarkaSen, I feel like, for $\Delta$ complexes, computing cellular homology is the same thing as computing simplicial homology. Am I wrong in that?
As for that $\Bbb{CP}^n$ problem…
> Compute homology of $\Bbb{CP}^n$, which is defined by gluing even dimensional cells upto dimension $2n$, using some wacky attaching maps $S^{2n-1} \to \Bbb{CP}^{n-1}$ inductively. (Hint: do you really need to know what the attaching map actually does?). Can you do it with $\Bbb{RP}^n$, which is slightly harder?
…I haven't actually done that yet; I'll get back to you
but I suspect it's going to lead naturally into the ideas of cellular homology.
I could answer your question but why bother to restrict yourself to singular homology groups of a space when you could define the homology category of a topos? :)
The definition of simplicial homology is much cleaner and more straightforward than cellular homology, making the ideas more obvious, and it's much easier to see that there might be relationship between simplicial homology and singular homology than it is with cellular, which looks completely different
Cellular homology, in some sense, is a calculational tool (in my mind), while simplicial homology is an object of interest
in any case, you're right that you could define homology in a number of equivalent ways using different combinatorial decompositions of your space, but simplicial homology is straightforward and comprehensible, and the broadest thing that's simultaneously a simplification and generalization is to just go ahead and do cellular homology, where the geometric intuition is less obvious, and sometimes less useful
try to understand poincare duality via cellular homology
Taking the pair $(\Bbb{CP}^n,\Bbb{CP}^{n-1})$, and taking advantage of the fact that $\Bbb{CP}^n/\Bbb{CP}^{n-1}=S^{2n}$, we get that they have the same homology for dimensions other than $2n+1$ and $2n$ (EDIT: typos)
That's the zero map, because the generator of the left thing is $2n$-dimensional, so its boundary would be $2n-1$-dimensional, but $\Bbb{CP}^{n-1}$ is only $2n-2$-dimensional!
Or, wait. I can do that cleaner.
I already know that $H_k(\Bbb{CP}^{n-1})=H_k(\Bbb{CP}^{n-2})$ for $k\ne 2(n-1),2(n-1)-1$
So $H_{2n-1}$ of it is just $H_{2n-1}(\Bbb{CP^0})=0$
Cellular chain complex (which you may not read up from Hatcher or somewhere else) arranges these groups in a chain complex. That's the cellular chain complex, computing the homology of which (funny how we are computing homology of homology groups, no?) gives you a homology theory for CW-complex which can e computed just out of the cell structure.
In fact that homology theory is isomorphic to the singular homology theory. So it's also independent of the cell structure you choose.
@robjohn: I got deserved downvote(s) on this. But I don't think anyone can actually answer the question. Do you have any thoughts? I'm quite interested.
@MikeM: That bottle of gin should help both of us.
What's gonna happen, we think, is my math teacher is gonna teach me real analysis, and then we'll continue into next year (junior HS) and when we're done with that we'll do complex analysis
And then we're gonna run into problems because the guy teaching my math is a physics professor so he might not be qualified to teach me much beyond that.
I tried to make editors sign a contract saying they'd keep prices down, but they don't hold to it. Actually, that book's been around over 10 years now.
If you can't find a cheap copy or afford it, DogAteMy, seriously, email me.
Plus, both my legs are rotated outwards a lot, which is bad. They fixed my right left during the surgery but I think they're gonna fix my left in August
I had a back brace for my scoliosis for a while. (Up until my surgery, in fact.) My doctor warned it might not work, because my back was pretty bad, but apparently it surprised him. Also, he said I recovered remarkably well from the leg-lengthening procedure. Like, he did it to hundreds of patients, and I recovered the best out of all of them
Do you think that all math undergraduates should take a course in topology? Topology is not required at my university (some treatment is given through analysis of course). But I've heard that topology is one of the pillars of a math foundation (along with analysis and algebra)
@rorty I took topology (as an elective) as an undergrad. It's a challenging course, but imo is very worth taking. So much of what you will do in adv. Calculus and Real Analysis and Differential Geometry even will makes a lot more sense and be easier to grok if you have the background in topology.
It turns out, for example, that many "canonical" constructions and concepts of topology can be abstracted and applied to other kinds of structure-giving an "alternative" kind of structure to investigate besides just "algebraic".
i have a piece wise linear function that is strictly increasing in each piece, and i was wondering if there was a good way to approximate it with a continuous function that is also strictly increasing with error of at most ~1/500 at each node
The concept of limits and continuity can be extended to other categories besides $\mathbf{Top}$ and its cousins..one can generalize topological spaces to "sets with structure"-said structure does not have to consist of operations in the sense of universal algebra, and you can have "hybrid structures" such as vector lattices (like the vector lattice of step-functions).
we can define the n-skeleta as $\displaystyle X^n = X^{n-1} \cup_{\phi} \bigsqcup_{\beta \in B} E^n_{\beta}$ and does it follow that a CW complex $X$ has $X^n$ a quotient space of $X^{n-1} \bigsqcup_{\beta} E^n_{\beta}$?
@AkivaWeinberger i have a set of unevenly spaced integer points from 0 to 20 that i have to round a set of rational points in that interval
say there are 7 points that i have to round to, i thought it would be fun if i could make a function that maps my points from 0 to 20 onto 0 to 7 (both floating point) so that i just have to run a simple round function on the result
@MikeMiller So we're looking for a smooth map $f : M \to N$ of smooth manifolds so that there is a vector field $X$ on $M$ (which is a choice of a tangent vector $X_p \in T_p M$ at each point $p$ which is smooth in the sense that $p \mapsto X_p$ defines a smooth map $M \to TM$) such that the map $f(M) \to TN$ defined by $f(p) \mapsto df_p(X_p)$ is not smooth? But that doesn't make sense because $f(M)$ (e.g., figure eight) need not be a manifold. So not sure if I understand the question.
Namely, take the immersion $f : S^1 \to \Bbb R^2$ which takes $S^1$ to figure eight and $p, q$ be points on $S^1$ which map to the bad point in the image 8.
$df$ takes different values at $X_p$ and $X_q$, not?
