Topology Homework 1: Function f and Homeomorphism Classes, Exercises of Designs and Groups

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

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Allison Arnold-Roksandich
Math 531: Topology
Homework 1
October 8, 2014
1Consider the function f: Int(B2)R2given by f(x) = 1
1−||x| | x.
(a) Show that fis 1-1.
(b) Show that fis onto and find an inverse function gto f.
(a) Suppose that f(x) = f(y). Then, we have that
1
1 ||x| | x=1
1 ||y|| y.
If we consider x=hrx,θxiand y=hry,θyias the polar coordinate forms, we then
get that
1
1 ||x| | x=1
1 ||y|| y
1
1rx
hrx,θxi=1
1ry
hry,θyi
rx
1rx
,θx=ry
1ry
,θy.
Thus we have that θx=θy. Furthermore, we get that
rx
1rx
=ry
1ry
rxrxry=ryryrx
rx=ry.
Therefore, x=y, and in conclusion, fis injective.
(b) Let yR2. Consider when x=1
1+||y|| y, which is in Int(B2)since ||y||
1+||y|| <1. Then,
f(x) = f1
1+||y|| y
=1
1
1
1+||y|| y
1
1+||y|| y
=1
1+||y||−||y||
1+||y|| 1
1+||y|| y
=y.
1
pf3

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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.