7/28/2019 math148ps1 (Algebraic Topology Problem Set 1)

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Math 148 Problem Set 1 Imran ThobaniFebruary 15, 2013

3) c. If f is continuous, then for each F F, whenever V Y is closed, f1(V) is closed,so by Theorem 17.2 of Munkres, F f1(V) = f|1F (V) is closed in F, so f|F is continuousfor each F F. If f|F is continuous for each F F, then whenever V Y is closed, foreach F F, f|1F (V) = F f1(V) is closed in F. Then F \ (F f1(V)) is open in F,which implies F \ (F f1(V)) = F O for some open O, so taking the complement in Fon both sides, gives that F f1(V) = F O where O is closed in X, and since F is alsoclosed in X, F f1(V) is closed in X. Then FF(F f1((V)) = f1(V) is closed in X(here we use the fact that Fis finite to ensure the union is closed), so f is continuous.

d. (X [0, 1]) = X [0, 1], which is an element of the basis of the product topology, sinceX is open in X and [0, 1]

is open in [0, 2], thus X [0, 1] is closed, and similarly, X [1, 2]is closed. It follows from part c. that H is continuous.

5) Take non-empty open U Rn, then U can be written as a union of open intervals in Rn(this is because the set of open intervals in Rn forms a basis for the topology on Rn). Takeany one of these intervals (c1, d1) ... (cn, dn); for each i, 1 i n, there exists qi Qsuch that ci < qi < di, so (q1,...,qn) Qn U. This proves that Qn is dense in Rn.

6) c. We know that x and y are continuous functions, and since the sum of continuousfunctions and product of continuous functions is continuous, it follows that a polynomialfunction is continuous. Also,

x is a continuous function, and the composition of continuous

functions is continuous, so the square root of a polynomial is continuous. From these facts,it follows that fi is continuous for i = 1, 2, 3. Then it follows by the universal property ofthe product topology that f(x, y) = (f1, f2, f3) is a continuous function.

Suppose (x, y) S1. Then x2+y2 = 1 so f(x, y) = (2x2+2y21, 2x

1 x2 y2, 2y

1 x2 y2) =(1, 0, 0). So f(S1) = {(1, 0, 0)}, i.e. if we let be the equivalence relation correspondingto D2/S1, then for a, b D2, a b = a = b a, b S1 = f(a) = f(b). Thenby the universal property for the quotient topology, there exists a unique continuous mapg : D2/S1 S2 such that f = g q where q is the quotient map from D2 D2/S1. Wehave g({[S1]}) = g(q(S1)) = f(S1) = {(1, 0, 0)}, i.e. g([S1]) = (1, 0, 0).

Take any (x,y,z) S2. Since f is surjective, there exists (a, b) D2 such that g(q(a, b)) =f(a, b) = (x,y,z), which shows that g is surjective. Suppose that g(x) = g(y). If x = [S1],then g(y) = g([S1]) = (1, 0, 0) so y = [S1] (otherwise, we would have y = [S1] and g(y) =g(q(q1(y))) = f(q1(y)) = f(y) where y = (y1, y2), y

2

1+ y2

2< 1 so f1(y) < 2y

2

1+ 2y2

2 1 < 1

so g(y) = f(y) = (1, 0, 0), a contradiction). So y = x.

By the same reasoning y = [S1] = x = [S1], so if, on the other hand, x = [S1], wemust have y = [S1], in which case we can write y = (y1, y2), x = (x1, x2) where x21 + x22