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## Main Question or Discussion Point

Are there any good online introductions to topology?

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Are there any good online introductions to topology?

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JasonRox

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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.

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JasonRox

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That's not an introduction to topology!

Not even close.

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JasonRox

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http://www.pdmi.ras.ru/~olegviro/topoman.html

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

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mathwonk

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

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mathwonk

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

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What prerequisite math is needed for topology?

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mathwonk

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probably calculus.

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

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JasonRox

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I did Topology before Analysis and I found it to be fine, in fact probably better. Of course, Metric Spaces is easier after Analysis and so on.It would be a little rough doing topology without first doing some proof based calculus.

Otherwise, I don't think there is any pre-requisites to Topology.

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mathwonk

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mathwonk

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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

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mathwonk

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First Concepts of Topology

Chinn, W. G. & Steenrod, N.e.

Bookseller: aridium internet books

(Cranbrook, BC, Canada) Price: US$ 7.90

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Quantity: 1 Shipping within Canada:

US$ 8.96

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Book Description: SInger, 1966. Trade Paperback. Book Condition: VG. First Printing edition. Usual library markings in and out. non-circulating. very light use, clean crisp pages. edge rub/wear. A solid copy.; Ex-Library. Bookseller Inventory # 10917

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

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mathwonk

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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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mathwonk

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Limitations on the method:

I am tempted to call these "chicken little" arguments, i.e. we prove "either there is a solution or else the sky would fall, and anyone can see the sky is not falling." In this case the "sky falling" is the statement the reals are not complete.

I.e. how close are we to actually finding a solution of f(x) = 0 by the IVT argument? We know there is a solution between a and b, but where? Well we could subdivide the interval further, into subintervals of length 1/10n, say. If we evaluate f at every endpoint of these subintervals, we can reapply the argument and deduce that there is a solution, not necessarily the ¬ in the proof, in one of these subintervals, assuming it is actually practical to evaluate f at these points. Thus in a finite amount of time, maybe large, we can find a number that is within any desired distance of a solution. I.e. we can find for each n, a point that agrees with a solution up to the first n decimal places.

How close is such an approximation to being a solution? There are two ways to measure how close µ is to being a solution of f(x) = 0. We could ask that |L-µ| be small, where L is an actual solution, or we could ask that f(µ) be close to zero. These are somewhat different. I.e. given any N, and any L in [a,b], it is possible to construct a continuous function f on [a,b] which equals -1,000 at every point x <= L -1/10N, and equals 1,000 at every point x >= L + 1/10N. Thus if N is very large, and if in the finite amount of time we allot to the problem, we do not choose x close enough to L, we might always have |f(x)| >= 1,000, for all "approximate solutions" of f(x) = 0.

Improvements on the method

What we really want in an approximate solution of f(x) = 0, is a point x which is near a solution, and also with f(x) close to zero. For this it would help to know that f is differentiable and to have a bound on the derivative of f. I.e. if we knew the derivative of f was never larger than a certain bound, then by the MVT we could bound how fast f grows, and thus know how fine to make our subdivision to get a good approximate solution, i.e. one with f(x) small. E.g. if we know the derivative of f is never greater than 10M, then on a subinterval interval of length 1/10N+M it cannot grow more than 1/10N. Thus for any subinterval with f negative at one end and positive at the other, all points x of this subinterval will be close to an actual solution, and will also satisfy |f(x)| <= 1/10N. Thus differential calculus is a big help in studying these questions.

Moral:

Are these computational limitations so serious as to make our use of topological and differential tools pointless? I think not. It is true we are limited in the precision of the results we obtain, but without these existential tools we would be very hard pressed to say anything at all about such difficult questions as those we will attack. In fact it is exactly their lack of precision that gives these tools their increased flexibility and hence their power. If we are not happy with these imprecise results, fine, try and do better. If we cannot, accept what they give us in the meantime.

Here is an example from my own research interests in algebraic geometry. It was proved over 20 years ago, using a variation of the inverse function theorem, that most “Prym varieties” arise from a unique double cover of curves, but to this day almost no examples of Prym varieties which do in fact arise uniquely are known. This is the question of the number of solutions of the equation p(X) = P, where p is the “Prym map”.

To this day the much harder problem of making this result precise is still unsolved. One step toward this result, uses deformation theory (i.e. abstract differential calculus).

In this problem there is a good guess as to what the answer is, i.e. what the right collection of unique solutions X is. In that situation one can try to use differential topological methods to verify the guess. In this way one can sometimes use these tools to get precise results too. For example suppose one wishes to show that a map from the circle to the plane is everywhere injective. If one can show the map has injective derivative everywhere, that the range is contained in the circle, and that there is at least one point of the range with only one preimage, then I claim that all points of the circle have exactly one preimage. Can you see intuitively why?

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mathwonk

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Generalizations

Now if we really understand the IVT argument above, shouldn't we be able to generalize it? Suppose we consider a complex polynomial

f:C-->C and ask whether it has a zero. This time restricting to an interval is not so useful since we cannot conclude that the image curve in C passes through zero just from knowing its endpoints are on opposite sides of zero. Try to understand the structure of the previous argument, i.e. the topology. We had an interval [a,b], and we deduced the existence of a solution in the interval just by looking at the function restricted to the endpoints {a,b}. What would be an analog of that in two dimensions?

We could look at our polynomial f on a square in C, or perhaps a disc. In order to deduce the existence of a root of f(x) = 0 lying in the disc, what would we look at? As an analog of the endpoints of the interval perhaps we should look at the restriction of f to the boundary of the disc. What behavior of f on the boundary would lead us to conclude there is a solution inside the disc? Let us take a very simple case. What if f were the identity on the boundary of the unit disc? Would it follow that f(x) = 0 has a solution inside the disc?

This is a direct analog of the IVT as follows. Consider this question: Is there a continuous map g:[a,b]--> {a,b}, from the interval [a,b] to its set of endpoints {a,b}, which maps a to a, and b to b? If there were a continuous map f:[a,b]-->R with f(a) < 0 and f(b) > 0, but with no point mapping to zero, then we could find a map g as above in the following way. Since the map R- {0}--> {a,b} sending negative numbers to a and positive numbers to b is continuous, if we compose f with this map we would get a map g:[a,b]--> {a,b}, with g(a) = a and g(b) = b.

Thus the question whether there is a continuous map f:[a,b]-->R-{0} with f(a) < 0, and f(b) > 0, is equivalent to the question of whether there is a continuous map g:[-1,1]--> {-1,1} with f(-1) = -1, and f(1) = 1. This form of the question can be generalized to higher dimensions.

Conjecture: suppose f:C-->C is continuous and that the restriction of f to the unit circle {z: |z|*= 1} is the identity map. Then we claim f(z) = 0 has at least one solution for |z| < 1.

How would we prove this? Remember the philosophy is to deduce a catastrophe from the falsity of the result, so assume it is false. I.e. that f maps the whole unit disc into C-{0}. Then what could one do? By analogy with the final form above of the IVT, after composing f with a retraction onto the circle, via z --> z/|z|, one could deduce there is a continuous map of the disc to its boundary circle which restricts to the identity on the boundary. Is this possible? Why or why not?

Retraction problem: There is no continuous map D-->?D of the disc onto its boundary, which restricts to the identity map on the boundary.

This is the hard result, analogous to the IVT above. It already implies the conjecture above that a polynomial f:C-->C which restricts to the identity on the unit circle has a root inside that circle. We can also use it to easily prove the Brouwer fixed point theorem.

Fixed point theorem: Any continuous map f:D-->D from the disc to itself has a fixed point.

proof: Assume not. Then for every x in D, f(x) and x are different points, hence determine a vector. Running along this vector from f(x) to x and then on until we reach the boundary gives a continuous map from D-->?D, which is the identity on the boundary, contradicting the retraction principle. QED.

Counting roots

What about results on the number of roots? I.e. if a complex polynomial restricts to the identity on the unit circle, how many roots do you think it should have inside that circle? How does the derivative of a complex polynomial behave? We will need this to generalize the methods above. I.e. recall the derivative of a complex polynomial at the point a, is a complex linear transformation L:C-->C.

If L is non zero, it is a complex linear isomorphism, in particular it is an orientation preserving injection. Thus if x is a root of f which is not a critical point, then near x, by the inverse function theorem f itself is also an orientation preserving injection. In particular f maps a small oriented disc centered at a, onto a small similarly oriented (deformed) disc centered at 0. If f restricts to the identity on the unit circle then how many roots do you think f can have inside the circle? Why?

Do you know how f behaves near a point where the derivative is zero? Does this suggest to you whether the derivative of an f can be zero inside the unit circle, if f itself is the identity on the unit circle? What is the key concept involved here, and how can we measure it? (Winding number. Sum of interior winding numbers at all roots, equals winding number of boundary.)

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mathwonk

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( the curly d boundary symbol seems to be a question amrk)

What next?

How would we generalize these ideas to still higher dimensions? Deduce a fixed point theorem in three dimensions from an appropriate retraction principle. How do we prove these retraction principles?

Retraction problem:

Let B be the unit ball in Rn, and f:B-->?B a continuous map. Prove that f cannot restrict to the identity on ?B.

I know two approaches to this problem. One uses the concept of winding or “wrapping” number. We associate to each map of a sphere to Rn-{0} the winding number of the image around 0. Then one views the ball B as a family of spheres, one for each value of the radius between 0 and 1. On one hand the winding number is a continuous function of the radius. But for r = 1 the winding number is one by hypothesis, and for r = 0, the winding number is zero. since the sphere of radius zero is only one point. This contradicts the IVT. One way to give a precise version of this argument is to use Stokes theorem to give a careful treatment of the notion of winding number.

The other approach is to study the geometry of the inverse image of one point. Since ?B has dimension one less than B, the inverse image

f-1(y) of a point is one dimensional near any point x where the derivative of f is surjective. If the derivative is surjective at every point of B of f-1(y), then ask yourself what f-1(y) looks like. The set f-1(y) is compact and at every point x of B-?B it looks (by the implicit function theorem) differentiably like an open interval. Since the only point of ?B in the inverse image is y itself, f-1(y) is thus a compact, smooth curve with only one endpoint. This is impossible. To use this approach requires the proof of the existence of a point y with f'(x) surjective for every x with f(x) = y. This depends on Sard's theorem. It also requires a careful description of all smooth compact curves.

The book for our course prefers the second approach, via the inverse function theorem and Sard's theorem, as more intuitive and simpler, and uses it in the first 3 chapters. The other approach is used in the last chapter. Intuition is very subjective however and I myself believe the first approach equally intuitive or moreso, at least for the problem as stated. Stokes theorem is also easier technically than Sard's theorem, but we will follow the book's lead as geometric and beautiful.

The book's approach also works in greater generality than the situation described above. To generalize the winding number approach to manifolds more general than balls does require a bit more work. In two dimensions however there is an easy approach to the winding number method in a disc that does not even require Stokes theorem. We will give it as an exercise in compactness.

We want to show the identity map of the circle to itself cannot extend to a continuous map of the disc to the circle. There are two steps.

I. A continuous map of the disc D to the circle ?D factors (by a continuous map) through the polar coordinate map R-->?D, taking t to (cos(t),sin(t)).

II. There is no injective continuous map ?D-->R.

Corollary: There is no continuous map f:D-->?D that restricts to an injection on ?D.

proof: Any continuous map f:D-->?D factors through a map D-->R whose restriction to ?D is not injective, hence the restriction of f to ?D cannot be injective either. QED.

Exercise: Prove I and II above. II is easier, since it follows from standard theorems of one variable calculus. I requires more knowledge of compactness, such as the concept of Lebesgue number of a covering. You might begin by showing the special case of II that there is no continuous lift of the identity map ?D-->?D through the polar coordinate map. Trying to construct one is very instructive for proving I.

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mathwonk

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Solution to Exercise: We do the easier one first.

Proof of II: Since ?D is compact (or if you want to remain within the realm of usual one variable calculus, just note that ?D can be parametrized by a closed bounded interval using the polar coordinate map), any continuous map g:?D-->R has an absolute maximum, say at p on ?D. Then if we regard a small arc around p as an interval using the polar coordinate map, we have a continuous map g:[p-e,p+e] -->R which has a maximum at p. Then both g(p-e) and g(p+e) are <= g(p), and if g is injective then both g(p-e) and g(p+e) are < g(p). By the intermediate value theorem, then g is not injective on the interval [p-e,p+e], a contradiction. QED.

Warm-up to I:

Before proving I we take the hint and try to construct a continuous factorization of the identity map ?D-->?D through the polar coordinate map.

First we replace the identity map ?D-->?D, by a parametrization

L:[a,b]-->?D of the circle by a closed bounded interval which is injective except that L(a) = L(b). (The polar coordinate map will work here too of course.) Then our problem becomes one of factoring L through the polar coordinate map by a continuous map T:[a,b]-->R such that T(a) = T(b). I.e. such a map will induce a factorization of the identity map ?D-->?D.

To try to produce the factorization T is easy except for the last condition. I.e. choose any appropriate value of T(a), i.e. any choice of polar coordinates for the point L(a). Then we want to extend T continuously to a nbhd of a. A small nbhd of a maps by L into the half circle containing L(a). The polar coordinates of points on this half circle form a disjoint collection of open intervals in R, only one of which contains the already chosen point T(a). Thus we must choose the factorization T on this whole interval about a to have values also lying in that interval of R. Thus there is at most one way to extend T to this nbhd of a. Moreover each of those open intervals in R is mapped by the polar coordinate map homeomorphically (i.e. continuously invertibly) onto the half circle about L(a). Hence inverting the polar coordinate map on any one of them is uniquely possible. Thus there is exactly one way to extend T to this nbhd of a.

Now each point of the circle lies in an open half circle, and since [a,b] is compact, we can choose a Lebesgue number for the corresponding open cover of [a,b] obtained by pulling this cover back by L. I.e. we can subdivide [a,b] into a finite number of subintervals such that each subinterval maps wholly into one half circle of ?D. Thus having once chosen T(a), we can extend T uniquely continuously to [a,b], one subinterval at a time, as above.

We remark that the factorization of T just found cannot have T(a) = T(b), since the injectivity of L forces the injectivity of T on [a,b). Thus [T(a),T(b)) is a half open interval of R and hence T(a) cannot equal T(b).

End of warm-up.

Proof of I: We want to construct a similar factorization of a continuous map D-->?D. To simplify the construction we assume we have a continuous map L:S-->?D, where S is a rectangle. Then choose again a Lebesgue number for the open cover of ?D by half circles, and subdivide S so fine that any closed sub rectangle of the partition of S map by L into a single half circle of ?D.

Then we can define T on the upper left corner p of S to be any appropriate polar coordinate of L(p). Now moving along the upper row of sub rectangles we extend T uniquely continuously as before. Then when we reach the second row of sub rectangles, we still can extend continuously by our choice of the Lebesgue number. I.e. each new sub rectangle U shares an edge with at most two rectangles on the previous row, and these 2 edges* being connected, are both mapped by T into the same interval of polar coordinates parametrizing some half circle containing L(U).

I.e. look at L(U) and find a half circle containing it. Then consider the disjoint collection of intervals of R providing polar coordinates for this half circle. Exactly one of them contains the connected image of the edges U shares with the sub rectangles where T is already defined. Thus using this interval we uniquely extend T to U, agreeing with the previous definition of T on those edges. This gives eventually a continuous extension of T to all of S, which is uniquely determined by the choice of T(p).

A similar subdivision argument works on a disc D, by inscribing D in a rectangle S and subdividing S as above. QED.

Remark: The argument used here contains the idea of "covering space". I.e. the polar coordinate map is a typical example of a covering space map, a map which always admits factorizations of maps defined on simply connected spaces.

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mathwonk

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JasonRox

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Even for Munkres I doubt you need any pre-requisites.

I say as long as you seen some Calculus and like know what a proof is then it's fine. Seems like as long as you read it carefully it's alright.

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mathwonk

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