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Let B be the closed ball in $\Bbb R^2$ with centre at the origin and radius unity. Pick out the true statements.

(a) There exists a continuous function $f : B \to\Bbb R$ which is one-one.
(b) There exists a continuous function $f : B \to\Bbb R$ which is onto.
(c) There exists a continuous function $f : B \to\Bbb R$ which is one-one and onto.

continuous image of a connected (compact) space is connected (compact). but how can i use this result in this problem

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I assume here that by R you mean $\mathbb{R}$ the set of real numbers. First notice that $\mathbb{R}$ is unbounded and thus not compact. You know that $B$ is compact. This is enough to handle the onto statement.

Let us consider the restriction of $f$ to the unit circle $S$. Since $f(S)$ must be compact and connected, it must be of the form $[a,b] \in \mathbb{R}$. If $f$ is 1-to-1 then $a \not= b$. Next, observe that if we remove any one of the uncountably infinite number of points from $S$ then we get a space that is homeomorphic to $\mathbb{R}$ and thus connected. On the other hand, there are only two points that we can remove from $[a,b]$ that will result in a connected space, the two points being $a$ and $b$. So, for any point $x \in (a,b)$, there are two distinct points in $S$ which gets mapped to $x$ . This should be enough to handle the 1-to-1 statement.

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