# $f:S^1\rightarrow S^1$ injective but not surjective

I thought this question was trivial, but I actually can't answer it, I hope I'm not missing something important.

Let $S^1:=\{z\in \mathbb{C} \textit{ such that } |z|=1\}$.

Can there be an injective continuous function $f:S^1\rightarrow S^1$ which is not surjective?

Then this question generalizes to:

Consider $M_n$ a n-dimensional differential compact manifold. Can there be $f:M_n\rightarrow M_n$ continuous and injective but not surjective?

• I hope you mean a continuous, injective function? Also, some times it definitely works, because you can map $\Bbb R^n\to \Bbb R^n$ by $(x_1, x_2, \ldots,x_n)\mapsto (\arctan(x_1), x_2,\ldots,x_n)$. Perhaps you want to limit this to compact manifolds? Then I suggest you look at the concept of the degree of a continuous function. Jun 13, 2017 at 10:59
• to find a counterexample, try a function $f:S^1\to S^1$ that is not continuous. Jun 13, 2017 at 11:04
• @Arthur yes, I mean continuous and injective, I corrected it Jun 13, 2017 at 11:10
• Would this still be not true if $f$ is just injective and not continuous? Jun 13, 2017 at 11:22

It's not possible, an injection would define an homeomorphism on its image (here I use compactness of $S^1$), and there is no subset of $S^1$ homeomorphic to $S^1$.

• Thank you. This argument should be valid also for any closed oriented surface, right? Jun 13, 2017 at 11:13
• Sure, I even think with homology you can prove it for any compact oriented manifolds.
– user171326
Jun 13, 2017 at 11:14
• For functions which are only injective (not continuous) this would still be true? Jun 13, 2017 at 11:17
• @user294185 : no, consider the map $f : S^1 \to S^1, e^{i\theta} \mapsto e^{i\theta/2}$, $\theta \in [0, 2\pi)$.
– user171326
Jun 13, 2017 at 11:29
• so this would not be true, there is the function you showed which is injective and not continuous and not surjective, am I right? Jun 13, 2017 at 11:33

I think the answer is no. Suppose $f:S^1 \to S^1$ is continuous injective but not surjective. Then the image of $f$ is a connected compact subset of $S^1$ that's properly contained in $S^1$. In particular, its image is homeomorphic to a closed interval. But then $f$ is a continuous bijection from a compact space to a Hausdorff space, and hence a homeomorphism, a contradiction.

In general, however, it's possible to find such functions. Take the real line for example. A homeomorphism to an open interval should do the trick.

• Ah oops, didn't see the other answer Jun 13, 2017 at 11:18
• it happens, +1 for adding more details than my answer :)
– user171326
Jun 13, 2017 at 11:33

For a compact orientable $n$-dimensional manifold $M$ you need to use the following two facts: (1) the homology group $H_n(M)$ is nontrivial; (2) the homology group $H_n(M\setminus\{pt\})=0$ where $pt$ is a single point. Therefore there are no injective nonsurjective continuous maps from $M$ to itself.