Let I denote the unit interval $[0, 1].$ Which of the following statements are true? 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
 A: 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.
A: (a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $B\setminus \{0\}$ is connected and $f[B\setminus \{0\}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $\mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
