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Bobby Moretti Math 441 Homework # 11/07/
Munkres, §18, problem 8 Let Y be an ordered set in the order topology. Let f , g : X → Y be continuous. (a) Show that the set { x | f (x) ≤ g(x) } is closed in X. (b) Let h : X → Y be the function
h(x) = min { f (x), g(x) }.
Show that h is continuous. [Hint: Use the pasting lemma.]
(a) Let B = X r { x | f (x) ≤ g(x) }. If B is empty, the result follows trivially, since then B is the empty set, which is open. So assume that B is nonempty, and fix any b ∈ B. Then we have b ∈ {/ x | f (x) ≤ g(x) }, so g(b) < f (b). Let m = minY , if it exists, otherwise let m = −∞. Similarly define M = maxY , or in the case that maxY does not exist, define M = +∞. Then define
Ub =
m, w
if there exists an element w ∈
g(b), f (b)
m, f (b)
otherwise,
and likewise define
Vb =
w, M
if there exists an element w ∈
g(b), f (b)
g(b), M
otherwise.
These are both nonempty open intervals (or rays) under the order topology Since f is continuous, the preimage of Ub under f , described by
f −^1 (Ub) = { x ∈ X | f (x) ∈ Ub } = { x ∈ X | f (x) < f (b) }
must be open in X. And similarly, the preimage of Ub under g, described by
g−^1 (Vb) = { x ∈ X | g(x) ∈ Vb } = { x ∈ X | g(b) < g(x) }
must also be open in X. Consider their intersection
Wb = f −^1 (Ub) ∩ g−^1 (Vb) = { x ∈ X | g(b) < g(x) < f (x) < f (b) }.
This set Wb is open as well, since it is the intersection of two open sets, and further, it is a neighborhood of b. So consider the union
W =
⋃
b∈B
Wb.
We claim that W = B. For if w ∈ W , then w ∈ Wb for some b. Then we have
g(b) < g(w) < f (w) < f (b).
But this gives us the condition that g(w) < f (w), and hence w ∈ B. We have proved the inclusion W ⊆ B. Now on the other hand, if b ∈ B, then b ∈ Wb ⊆ W , since each set Wb is a neighborhood of b. So B ⊆ W. So B is a union of open sets, and hence is itself open. Thus the set X r B = { x | f (x) ≤ g(x) } is open. (b) Let A = { x ∈ X | f (x) ≤ g(x) } , and let B = { x ∈ X | g(x) ≤ f (x) }. By part (a), we know that both A and B are closed sets, and it is clear that A ∪ B = X. The restrictions f|A : A → Y and g|B : B → Y are both continuous. For every x ∈ A ∩ B, we have x ∈ A and x ∈ B. Then f (x) ≤ g(x) and g(x) ≤ f (x), so f (x) = g(x) for all x ∈ A ∩ B. Since A and B are closed, A ∩ B is closed as well. So by the pasting lemma, we have that h is continuous.
Munkres, §18 problem 10 Let f : A → B and g : C → D be continuous functions. Let us define a map f × g : A ×C → B × D by the equation
( f × g)(a, c) = f (a) × g(c).
Show that f × g is continuous.
Let B be any basis open set in C × D. Then B = U ×V , where U is some open set in C and V is some open set in D. Its preimage is given by
( f × g)−^1 (U ×V ) = { (a, c) ∈ A ×C | ( f × g)(a, c) ∈ U ×V } = { (a, c) ∈ A ×C | f (a) ∈ U and g(c) ∈ V } = { a ∈ A | f (a) ∈ U } × { c ∈ C | g(c) ∈ V } = f −^1 (U) × g−^1 (V ).
Since f and g are continuous, the sets f −^1 (U) and f −^1 (V ) are open in A and C, respec- tively. So we have ( f × g)−^1 (U ×V ) = f −^1 (U) × g−^1 (V )
is the poduct of two open sets, and hence is open in A ×C.
Munkres §18 problem 11 Let F : X ×Y → Z. We say that F is continuous in each variable separately if for each y 0 in Y , the map h : X → Z defined by h(x) = F(x, y 0 ) is continuous, and for each x 0 in X, the map k : Y → Z is continuous. Show that if F is continuous, then F is continuous in each variable separately.
In other words, g−^1 (Ug) ∩ h−^1 (Uh) is a neighborhood of x. But x is a limit point of A, since x ∈/ A but x ∈ A. So every neighborhood of x intersects A at some other point. In particular, the neighborhood g−^1 (Ug) ∩ h−^1 (Uh)
contains some point a 6 = x such that a ∈ A. But then g(a) = f (a) = h(a). And certainly a ∈ g−^1 (Ug), so that h(a) = g(a) ∈ Ug. Also, we have a ∈ h−^1 (Uh), so that g(a) = h(a) ∈ Uh. Thus we conclude that
h(a) = h(a) ∈ Ug ∩Uh.
But this is impossible, since these sets are disjoint. We have arrived at a contradiction, so it cannot be the case that h 6 = g. Thus h = g is the unique continuous function extending f to A.
Munkres, §22 problem 2 (a) Let p : X → Y be a continuous map. Show that if there is a continuous map f : Y → X such that p ◦ f equals the identity map of Y , then p is a quotient map. (b) If A ⊆ X, a retraction of X onto A is a continuous map r : X → A such that r(a) = a for each a ∈ A. Show that a retraction is a quotient map.
(a) Let p and f be continuous maps as defined above. Suppose B ⊆ Y is a set whose preimage under p is open in X. Since the set p−^1 (B) is open in X, so must the set f −^1 (p−^1 (B)) be open in Y , since f is continuous. But then we have f −^1 (p−^1 (B)) = (p ◦ f )−^1 (B) = B since p ◦ f is the identity map, so B is open in Y. We have proved that if p−^1 (B) is open in X, then B is open in Y. The converse of this follows immediately, since p is continuous. Hence p is a quotient map. (b) Suppose B ⊆ A, and suppose that r−^1 (B) is open in X. But note that r−^1 (B) = { a ∈ A | r(a) ∈ B } = { a ∈ A | a ∈ B } = B. So B is open in A. Thus any subset B of A whose preimage is open is iteself open. The converse of this follows, since r is continuous. So r is a quotient map.
Munkres, §22 problem 4 (a) Define an equivalence relation on the plane X = R^2 as follows:
x 0 × y 0 ∼ x 1 × y 1 if x 0 + y^20 = x 1 + y^21.
Let X/∼ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set g(x, y) = x + y^2 .] (b) Repeat (a) for the equivalence relation
x 0 × y 0 ∼ x 1 × y 1 if x^20 + y^20 = x^21 + y^21.
(a) If (x 0 , y 0 ) is any point in R^2 , then its equivalence class is given by
[(x 0 , y 0 )] =
(x, y) ∈ R^2
∣ (^) x + y^2 = x 0 + y 02 }^ ,
which is a parabola symmetric about the x axis, opening to the left, with vertex at the point (x 0 + y 02 , 0 ). So X/ ∼ is the space of these parabolas. Now consider the map
g : R^2 → R
sending (x, y) 7 → (x + y^2 ). Now for any z ∈ R, certainly there exists some x 0 , y 0 ∈ R such that x 0 + y 02 = z. Then we have
g−^1 ({ z }) =
(x, y) ∈ R^2
∣ (^) x + y^2 = z
(x, y) ∈ R^2
∣ (^) x + y^2 = x 0 + y 02 } = [(x 0 , y 0 )].
Thus X/ ∼ =
g−^1 ({ z })
∣ (^) z ∈ R }. The function g is clearly continuous and surjective, and it is clearly an open map. So it is a quotient map. Hence g induces a homeomorphism between X/ ∼ and R. (b) This is the space of circles about the origin. The function g : R^2 → [ 0 , ∞) defined by g(x, y) = x 02 + y 02 is a continuous, surjective map that happens to be an open map. The fibers g−^1 ({ r }) happen to be circles of radius r, for r > 0, and the set { 0 }, when r = 0, which is precisely the set of equivalence classes in X/∼. So g induces a homeomorphism between X/∼ and the subspace [ 0 , ∞) of R.
