# 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

• why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-\{0\}$ is certainly connected. Jan 11, 2019 at 17:43
• @Randall B is a circle , cut the circle it will disconnect Jan 11, 2019 at 17:45
• I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected. Jan 11, 2019 at 17:46
• I don't know.... Jan 11, 2019 at 17:52
• @Randall compactness does help if you know dimension theory too: if $f: B \to \mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $\dim f[B] \le 1$ while $\dim B=2$. Bit overkill though. Jan 11, 2019 at 18:41

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) 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.