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Online introduction to topology. 
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#1
Mar1007, 08:16 PM

P: 140

Are there any good online introductions to topology?



#2
Mar1007, 08:40 PM

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PF Gold
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Personally, buying a cheap textbook at a second hard store is probably your best bet.
I bought mine for like $9 or so. I learned a lot. Although it wasn't as thorough as others, I thought it was perfect for a first timer. Perfect for an undergraduate anyways. When I went on to take Topology as a course, we had Munkres. I was able to answer questions without too much difficulty and so on, so I knew the cheap textbook taught me well. It felt good knowing I didn't waste my time reading it. The book I got, in case you're wondering, is by Theral H. Moore. I saw it at the University library as well, so I would imagine there exists enough copies around to get a hold of it. The questions weren't too hard. I say there were just right for a beginner. Honestly, I think they should teach out of this textbook as a Topology I course for like 2nd year students. 


#3
Mar1107, 04:06 AM

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#4
Mar1107, 05:50 AM

P: 53

Online introduction to topology.



#5
Mar1107, 01:46 PM

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PF Gold
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Not even close. 


#6
Mar1107, 01:48 PM

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PF Gold
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This book looks alright. It has the introductory ideas and what not. 


#7
Mar1107, 01:52 PM

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i think i hav e posted some introductroy notes before, maybe in the who wants to be thread?



#8
Mar1107, 02:05 PM

P: 140

Well keep trying! apparently, if I am to learn differnetial geometry properly i need some general topology.



#9
Mar1107, 08:28 PM

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you might strt with posts 115, 116 in the who wants to be a mathematician thread in academic advice.



#12
Mar1307, 06:19 PM

P: 617

It would be a little rough doing topology without first doing some proof based calculus.



#13
Mar1307, 06:41 PM

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Otherwise, I don't think there is any prerequisites to Topology. 


#14
Mar1307, 11:39 PM

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there is a high school level intro to topology by chinn and steenrod, with almost "no" prerequisites. it is excellent, chinn is a decorated high school teacher and steenrod is one of the most outstanding topologists of the 20th century, from princeton.



#15
Mar1307, 11:40 PM

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here is one from amzon.com for $20:
First Concepts of Topology (Paperback) by N. E. Steenrod (Author), W. G. Chinn (Author), William G. Chinn (Author) (2 customer reviews) Price: CDN$ 20.40 & eligible for FREE Super Saver Shipping on orders over CDN$ 39. Details Availability: Usually ships within 4 to 6 weeks. Ships from and sold by Amazon.ca. 6 used & new available from CDN$ 13.51 


#16
Mar1307, 11:42 PM

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heres an even cheaper one:
First Concepts of Topology Chinn, W. G. & Steenrod, N.e. Bookseller: aridium internet books (Cranbrook, BC, Canada) Price: US$ 7.90 [Convert Currency] Quantity: 1 Shipping within Canada: US$ 8.96 [Rates & Speeds] Book Description: SInger, 1966. Trade Paperback. Book Condition: VG. First Printing edition. Usual library markings in and out. noncirculating. very light use, clean crisp pages. edge rub/wear. A solid copy.; ExLibrary. Bookseller Inventory # 10917 


#17
Mar1407, 09:25 AM

P: 140

Would you consider it complete enough to learn differential geometry the "right " way?



#18
Mar1407, 11:46 AM

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If you were asking whetehr chinn and steenrod has akll one needs to elarn diff geom right, the answer is NO.
but it is a start. the right question is not does this book have everything i need, but does this book have something i need. to lesarn differential geometry one needs very little topology. maybe you should search the web for free topology books. there are not many but there are several with the sort of boring basic point set topology, out there which may be what you want. the more interesting and fun stuff involves some algebra, homology, homotopy, intersection theiory. after reading my posts in who wants to be a mathematician thread referred to above, here is more free introduction to ideas of topology: (but some msymbols like little curly d's, and maybe infinity symbols will not reproduce right) 4220, lecture 1, First steps, some problems and approaches to them. The goal of this course is to use calculus (i.e. the concepts of continuity and differentiability) to prove statements such as: a complex polynomial of positive degree always has roots, a smooth self mapping of the disc always has fixed points, or a smooth vector field on a sphere always has zeroes, and higher dimensional generalizations of them. These are called existence statements. They are called such because usually we do not produce the solutions whose existence is claimed, rather we deduce some contradiction from the assumption that no solution exists. Thus it is entirely another matter to obtain specific information about these solutions. I want you to give some thought in each case to the problem of actually finding, or at least approximating, these solutions. On the other hand we will often prove results about the "number" of such solutions. We use quotation marks because again there is no guarantee the actual number of solutions will obey our prediction. We will define at times a "weighting" for each solution, and will prove that either there are infinitely many solutions, or if the number of solutions is finite, the sum of the weights equals our number. Since a non solution has weight zero, it follows that if our predicted number is non zero, then there is at least one solution. Moreover if there is only one solution, then it must have weight equal to our predicted number. This is a big improvement, since in some cases we can actually find some of the solutions and their weights, and if their sum is deficient from our prediction, we then conclude there must be more solutions. This is a useful tool in plane algebraic geometry called the strong Bezout theorem. To take advantage of this, poses the challenge of actually computing these weights. In each case we encounter, please give some thought to how to calculate the weights, or the actual number of solutions. Let's recall one of the earliest cases of an existence theorem of this type, the so called "intermediate value theorem". Theorem (IVT): If f:[a,b]>R is continuous, and f(a) < 0 while f(b) > 0, then there is a number c with a < c < b and f(c) = 0. This proof is based on the completeness axiom for the real numbers: every non empty set of reals which is bounded above has a real least upper bound. proof of theorem: Consider the set S = {x in [a,b] such that f(x) <= 0}. S contains a, so S is non empty, and b is an upper bound for S, so S has a least upper bound L with a <= L <= b. If f(L) < 0 then L < b so f is also negative at some number between L and b, so L is not even an upper bound for S. If f(L) > 0, then L > a, so all members of S are less than L but there is an interval of numbers less than L where f is also positive, i.e. which are not elements of S. Hence every element of that interval is an upper bound of S, so L is not the least upper bound of S. This contradiction proves that f(L) is neither positive nor negative, hence must be zero. QED. abstract version of this proof: If f is continuous on D, and D is connected, then f(D) is also connected. In R the only connected sets are intervals, thus f([a,b]) is an interval. QED. Let's extend the argument a bit. Corollary: If f is a polynomial of odd degree, with real coefficients, then f has a real root. proof: Assume f is monic. Then the limit of f(x) is ? as x>?, and is ? as x > ?. Hence there exist a,b, with a < b and f(a) < 0 and f(b) > 0. QED. Corollary: If f is a differentiable function with no critical points in [a,b] and with f(a) < 0 < f(b), then f has exactly one root in [a,b]. proof: Uniqueness follows from the MVT. QED. Cor: If f is a polynomial of even (odd) degree with real coefficients, and no root is a critical point, then f has a finite even (odd) number of roots. proof: (even case) By the hypothesis that f'(x) != 0 when f(x) = 0, the graph of f crosses from one side to the other of the x axis at each root, but the limit of f(x) is ? both as x> ? and as x>  ?. So it crosses the x axis an even number of times, provided the number of crossings, i.e. roots, is finite. To show the number of roots is finite, note that if no root is a critical point, i.e. if f'(x) != 0 when f(x) = 0, then each root is isolated, i.e. each belongs to an open interval in which there are no other roots. Now since the limit of f(x) is ? both as x> ? and as x>  ?, we can choose a bounded closed interval [a,b] containing all the roots. Then by compactness of this interval an infinite set of roots would have an accumulation point. But such an accumulation point would be a non isolated root, contrary to hypothesis. QED. Note: This last corollary introduces and exploits the concepts of transversality and regular values. Notice too that it includes a version of the idea of the weight of a solution, since where the graph crosses the x axis the weight is +1 if f crosses in an increasing direction and 1 if f crosses in a decreasing direction. Then the theorem says the sum of the weights is one for odd degree monic polynomials and zero for even degree ones. We could also define the weight of a root where the graph stays on the same side of the x axis (and f' = 0) to be zero, provided we relax the assumption that roots are not critical points. (in this case however there can be infinitely many roots, so adding up their weights may not be possible.) Note that a root of weight zero can disappear if the graph is wiggled arbitrarily little, but not so for a root of non zero weight. This introduces the concept of "stability" of a root of non zero weight. Here as an easy corollary is a "fixed point" theorem. Cor: If f:[a,b]>[a,b] is continuous, then f has a fixed point, i.e. there is some point x in [a,b] with f(x) = x. proof: We try to translate the conclusion into that of the IVT, i.e. try to replace the existence of a fixed point by the existence of a zero. Consider g(x) = f(x)  x. Then f has a fixed point at x if and only if g(x) = 0. Now g:[a,b}> R is continuous and g(a) = f(a)a >= 0 since f(a) is in [a,b]. Also g(b) = f(b)b <= 0. Thus either a or b is a fixed point, or if neither is, then f has one between them, by the IVT. QED. Note: These corollaries are really easy and fun, as compared to the IVT which is hard and boring. Why then do we teach only the IVT in calculus courses? Lucky us, this course is about deducing the easy fun stuff from the hard boring stuff. (I admit I like the hard boring stuff too, so remind me to ease up if I begin to stress it too much.) 


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