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

  • 1
    $\begingroup$ why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-\{0\}$ is certainly connected. $\endgroup$
    – Randall
    Jan 11, 2019 at 17:43
  • $\begingroup$ @Randall B is a circle , cut the circle it will disconnect $\endgroup$
    – jasmine
    Jan 11, 2019 at 17:45
  • 2
    $\begingroup$ 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. $\endgroup$
    – Randall
    Jan 11, 2019 at 17:46
  • 1
    $\begingroup$ I don't know.... $\endgroup$
    – Randall
    Jan 11, 2019 at 17:52
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 11, 2019 at 18:41

2 Answers 2


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.


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