The existence of a 1-1 continuous map between two topological spaces.

Question

Show that there is no one-to-one continuous map $f$ from $\mathbb{R}^n$ to $\mathbb{R}^2$ for $n\gt 2$ with $f(0)=0$. I tried using the hint: consider $f:\mathbb{R}^n-\{0\}\rightarrow \mathbb{R}^2-\{0\}$ and the induced map: $f_*:\pi_1 (\mathbb{R}^n-\{0\}) \rightarrow \pi_1(\mathbb{R^2}-\{0\})$.

Answer 1

Since $f_*$ is zero you can lift $f$ to the universal cover of $\mathbb{R}^2-\{0\}$ which is $\pi: \mathbb R^2\to \mathbb{R}^2-\{0\}$, i.e. you have a continuous map $g:\mathbb{R}^n-\{0\}\to \mathbb R^2$ with $f=\pi\circ g$.
Since $f$ is injective, $g$ must be injective too.
But then the restriction $\gamma:S^{n-1} \to \mathbb{R^2}$ of $g$ to the sphere $S^{n-1}\subset \mathbb R^n-\{0\}$ would be injective too, which contradicts Borsuk-Ulam (after composing $\gamma$ with the embedding $\mathbb R^2\hookrightarrow \mathbb R^{n-1}$) .

Answer 2

Note that $\mathbb R^n\setminus\{0\}$ is simply connected if $n>2$ but not if $n=2$, and that says something about $\pi_1(\mathbb R^n\setminus\{0\})$.

Answer 3

On Georges' request, here is a detailed proof expanding Michael's incomplete argument.

Lemma 1. Let $(K_i)$ be an exhaustion of $R^n$ by a sequnce of compact subsets: $$ K_1\subset K_2\subset ... $$ Then $\exists i$ such that $K_i$ has nonempty interior.

Proof. This is an immediate corollary of Baire's theorem ($R^n$ cannot be a countable union of nowhere dense subsets). qed

Lemma 2. $X=R^n \setminus \{0\}$ is simply-connected if $n\ge 3$. Proof. $X$ is homeomorphic to $S^{n-1}\times R$, hence, homotopy-equivalent to $S^{n-1}$. To prove simple connectivity of the latter, note that every loop in it is homotopic to a piecewise-circular loop. The latter has nowehere dense image (as $n-1\ge 2$), hence, can be regarded as a loop in $R^{n-1}$. The latter is contractible. Hence, the original loop is null-homotopic. qed

(If you know Seifert - van Kampen theorem, it gives an alternative proof, but what I wrote is much easier than S-vK Theorem.)

Lemma 3. Let $f: R^n\to R^2$ be a continuous bijection. Then there exists a round circle $C\subset R^2$ such that the restriction $f^{-1}|_C$ is continuous.

Proof. Let $(B_i)$ denote an exhaustion of $R^n$ by closed round balls. Their images $K_i=f(B_i)$ are compact and, obviously, form an exhaustion of $R^2$. By Lemma 1, there exists $K_i$ with nonempty interior. Take a round circle $C$ contained in $K_i$. The map $f: B_i\to K_i$ is a continuous bijection; $B_i$ is compact and $K_i$ is Hausdorff. Hence, $f^{-1}|_{K_i}$ is continuous. Lemma follows. qed

Lemma 4. There does not exists a continuous bijection $f: R^n\to R^2$, $n\ge 3$.

Proof. Suppose that $f$ exists. Let $C$ be a circle as in Lemma 3. Let $p$ be the center of this circle and $r$ its radius. The map $g=f^{-1}|_C$ is continuous (Lemma 3). Then $g: C\to X=R^n \setminus \{q\}$ is also continuous, where $q=f^{-1}(p)$. Since $X$ is simply-connected (Lemma 2), the map $g$ extends to a continuous map $h$ of the disk $B(p,r)$ (centered and $p$ and of radius $r$). Composing $j=f\circ h$, we get a continuous map $j: B(p,r)\to R^2\setminus \{p\}$ which is the identity on $C$. Contradiction. (OK: here you need to know something about nontriviality of the 1st homology or fundamental group of the circle, but I assume that you already know this.) qed