

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Solutions to homework problems from a university-level topology course. The first problem involves showing that a given function is injective and surjective, and finding its inverse. The second problem deals with identifying homeomorphism classes of subsets in the plane based on their separation properties.
Typology: Exercises
1 / 3
This page cannot be seen from the preview
Don't miss anything!


Allison Arnold-Roksandich Math 531: Topology Homework 1 October 8, 2014
1 Consider the function f : Int(B^2 ) → R^2 given by f (x) = (^1) −||^1 x|| x.
(a) Show that f is 1-1.
(b) Show that f is onto and find an inverse function g to f.
(a) Suppose that f (x) = f (y). Then, we have that 1 1 − ||x||
x =
1 − ||y||
y.
If we consider x = 〈rx, θ x〉 and y = 〈ry, θ y〉 as the polar coordinate forms, we then get that 1 1 − ||x||
x =
1 − ||y||
y
1 1 − rx
〈rx, θ x〉 =
1 − ry
〈ry, θ y〉 〈 rx 1 − rx
, θ x
ry 1 − ry
, θ y
Thus we have that θ x = θ y. Furthermore, we get that rx 1 − rx
ry 1 − ry rx − rxry = ry − ryrx rx = ry.
Therefore, x = y, and in conclusion, f is injective.
(b) Let y ∈ R^2. Consider when x = (^1) +^1 ||y|| y, which is in Int(B^2 ) since (^1) +||y||||y|| < 1. Then,
f (x) = f
1 + ||y||
y
∥ (^1) +^1 ||y|| y
1 + ||y||
y
1 +||y||−||y|| 1 +||y||
1 + ||y||
y
= y.
Therefore, we get that f is surjective as we can find a value for x to produce any needed y. The inverse function for f is
f −^1 (x) = g(x) =
1 + ||x||
x.
2 For the ten subsets of the plan below, group into homeomorphism classes. Distin- guish each class by some specific property.
The first homeomorphism class is {(a)}. We can remove 0 and 1 with causing a separa- tion, but every other point causes a separation. The second is {(b)} where removing any single point will cause a separation. The third is {(c), (j)}. The point of intersection of the two circles is the only single point that when removed causes a separation. It is also possible to pick two points such that there is only one component after the removal. The fourth is {(d), ( f )}. There exists a point such that when it is removed we are left with four components. The fifth is {(e)}. There exists a point such that removing it creates three components. The sixth is {(g)}. There is no way to create a separation by removing only one point. The final homeomorphism class is {(h), (i)}. Both have an infinite number of points that when removed creates a separation, but not all points when removed create a separation.