Let $B := \{(x, y) \in \mathbb{R}^2
: x^2 + y^2 \le 1\}$be the closed ball in $\mathbb{R^2}$ with center at the origin.
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
Which of the following statements are true?
$(a)$ There exists a continuous function $f : B \rightarrow \mathbb{R}$ which is one-one
$(b)$ There exists a continuous function $f : B \rightarrow \mathbb{R}$ which is onto.
$(c)$ There exists a continuous function $f : B \rightarrow I × I$ which is one-one.
$(d)$ There exists a continuous function $f : B \rightarrow I × I$ which is onto.
I thinks none of option will be correct
option $a)$ and option $b)$ is false Just using the logics of compactness, that is $\mathbb{R}$ is not compacts
option c) and option d) is false just using the logic of connectedness that is $B-\{0\}$ is not connected but $I × I-\{0\}$ is connectedness
Is my logics is correct or not ?
Any hints/solution will be appreciated
thanks u
Best Answer
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I \times I$, as is well-known. A homeomorphism will fulfill both. Note that $B\setminus\{0\}$ is actually connected so your proposed argument doesn’t work.