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

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: $f_*:\pi_1 (\mathbb{R}^n-\{0\}) \rightarrow \pi_1(\mathbb{R^2}-\{0\})$.

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I tried that hint but it does not work. And next i tried Borsuk-Ulam theorem – Toeplitz Sep 1 '14 at 20:54
Are you required to prove it in a way of Alg Top? It would be fairly easy using only topological methods. – Anthony Peter Sep 1 '14 at 20:58
@AnthonyPeter It follows from dimension theory arguments, but that's hardly elementary for most. – Henno Brandsma Sep 1 '14 at 21:00
@AnthonyPeter Try to write such a thing down. You will likely have some significant difficulty. – Mike Miller Sep 1 '14 at 21:02
@MikeMiller I suppose you may be correct. Sometimes intuition may not respond well to practice – Anthony Peter Sep 1 '14 at 21:03

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 $\mathbb R^2\hookrightarrow \mathbb R^{n-1}$) .

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The conciseness of the answer should not conceal the fact that I have used two big theorems in algebraic topology: lifting condition for covering spaces and Borsuk-Ulam. One could also use invariance of domain (which is a hard nut too) but the sad principle of conservation of difficulty implies that no remark on the (non-) triviality of the fundamental groups involved will by itself solve the problem. – Georges Elencwajg Sep 1 '14 at 22:00
Dear Georges: With my comment to Hardy's answer, his proof goes through. This still leaves out a proof of simple connectivity of spheres of dimension at least 2, which requires very little work (PL approximation of continuous functions). – studiosus Sep 1 '14 at 22:07
Dear @studiosus, unfortunately I don't understand your comment. You (or Michael) are welcome to write a complete answer (please no supplementary comment), but until then I'm not be convinced (you may use simple connectedness of spheres without further explanation). If you or he do write such a complete answer, I'll be the first to upvote you and congratulate you. – Georges Elencwajg Sep 1 '14 at 22:22
Dear Georges: Done. – studiosus Sep 1 '14 at 22:52
Great post, @studiosus: I'll certainly study your answer, which looks very interesting. The OP actually asks about continuous injections, not bijections. Does your post also answer that question? Anyway, your result is very interesting in itself. – Georges Elencwajg Sep 2 '14 at 0:00

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

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Does $f$ one-to-one imply that $f_{\ast}$ cannot be $0$? – Henno Brandsma Sep 1 '14 at 21:01
@HennoBrandsma No: consider the embedding $\Bbb R \hookrightarrow S^1$. – Mike Miller Sep 1 '14 at 21:02
@MikeMiller I thought not, so one needs some extra argument then.. (I'm a general topologist, not an algebraic one, so my skills there are limited...) – Henno Brandsma Sep 1 '14 at 21:04
There is one ingredient missing in this argument: if you take, say, a circle in the plane, you need to know that the inverse of f is continuous in it. The way to do it is to exhaust the n-space by compact subsets and then use a standard fact of general topology, about continuius bijections from compacts to Hausdorff spaces. – studiosus Sep 1 '14 at 21:30
@studiosus : I expected that more than just one thing would be missing. I certainly didn't intend completeness this time. – Michael Hardy Sep 1 '14 at 22:24

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

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+1, as promised! – Georges Elencwajg Sep 2 '14 at 8:20