From the Plane to the Line

Relevant wiki: Connected Space

Suppose that the continuous function $f$ is injective. Let $f(0,0) = c$ . If we define $U \; = \; \big\{ x \in \mathbb{R}^2 \, \big| \, f(x) > c\big\} \hspace{1cm} V \; = \; \big\{ x \in \mathbb{R}^2 \,\big|\, f(x) < c \big\}$ then $U,V$ are disjoint open subsets of $\mathbb{R}^2$ whose union is $\mathbb{R}^2\backslash\{(0,0)\}$ .

If $U = \varnothing$ then $f(x) < c$ for all $x \neq (0,0)$ . Thus $f(1,0)\,,\,f(0,1) \,< \, c$ . If $f(1,0) < f(0,1)$ we can, using the intermediate value theorem , find $0 < u < 1$ such that $f(u,0) = f(0,1)$ , which is impossible. It is similarly impossible for $f(1,0) > f(0,1)$ . Thus it is impossible for $U$ to be empty. Similarly $V$ is not empty. But this tells us that $\mathbb{R}^2 \backslash\{(0,0)\}$ is not connected, which is false. Thus $f$ cannot be injective.

Relevant wiki: Borsuk Ulam Theorem

The restriction $g = f|_{S^1}$ of a continuous function $f$ to the unit circle is continuous.

Therefore, by the one-dimensional Borsuk-Ulam theorem, there are antipodal points $x,-x \in S^1$ such that $g(x) = g(-x)$ , so $f(x) = f(-x)$ . If $f$ were injective, then this would imply $x=-x$ , so $0 = x \in S^1$ , which is absurd, so $f$ cannot be injective.

Note that the one-dimensional instance of the Borsuk-Ulam theorem is basically the intermediate value theorem, but I've referenced it here since it provides a very nice generalization to higher dimensions.

I also want to point out that the dual form of the this question is also interesting, if not a fair bit harder. It is "Let $f : \mathbb{R} \to \mathbb{R}^2$ be continuous. Can $f$ be surjective?" [assuming the axiom of choice... otherwise, I believe it's undecidable]

Relevant wiki: Intermediate Value Theorem

Let's assume the contrary. Then, there exists a function $f$ which is injective.

Let $a = f(0,0), b = f(1, 0)$ . Without loss of generality, let's assume $a < b$ . By intermediate value theorem, $f$ restricted to $\left \{ (t,0) | 0 < t < 1\right \}$ assumes all values in the interval $(a,b)$ . In particular, $f$ assumes $\frac{a+b}{2}$ . Let $0 < \xi < 1$ be such that $f(\xi, 0) = \frac{a+b}{2}$ .

Now, let us look at $d = f(\xi,1)$ . By injectivity of $f$ , $d \notin [a,b]$ . Without loss of generality, let's assume $d > b$ . Then, similarly by looking at the restriction of $f$ to $\left \{ (\xi,t) | 0 < t < 1\right \}$ assumes all values in the interval $( \frac{a+b}{2}, d )$ . In particular, it assumes $b$ , which is a contradiction.

An interesting question:

What happens when we remove the condition that f is continuous?