No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$ A friend asked me if there could be a continuous injective map 
$$f: \mathbb{S}^1 \to \mathbb{R}.$$  
My intuition tells me no. Endow $\mathbb{S}^1$ with a topology $\mathscr{T}$ and fix a pole $x \in \mathbb{S}^1$, consider $f(x)$ and take $B_\epsilon (f(x))$. Since $f$ is continuous,  $f^{-1}(B_\epsilon (f(x)))$ is open so take $N(x) \in \mathcal{N}(x)$ so that $N(x) \subset f^{-1}(B_\epsilon (f(x)))$.  
Now I don't know enough topology to formalize my intuition. Namely, if $x_1$ is close on one "side" of $x$ and $x_2$ is close on the other side, then $f(x_1)$ is at most $\epsilon$ away from $f(x)$ and likewise with $f(x_2)$. So, either $f$ is constant or $f(x_2)$ approaches $f(x)$ as $x_2 \to x$, so that clearly there must be a point that is not injective.
So, how do I express this formally as a topological proof? 
 A: A hint:
No fiddling around with $\epsilon$'s here! You want to establish a certain global fact; therefore you have to bring in appropriate global tools, like the MVT, the fact that $S^1$ is compact, etc. 
Full solution, pedestrian way:
Since $S^1$ is compact the function $f$ assumes a minimum $a$ at a point $z_a\in S^1$ and a maximum $b>a$ at a point $z_b\in S^1$. These two points split $S^1$ into two arcs $\gamma_1$, $\gamma_2$. On each of the two arcs the function $f$ has to assume the value $m:={a+b\over2}$ somewhere, which contradicts injectivity.
A: Note that a continuous bijection from a compact set to a Hausdorff space is a homeomorphism. If such a map $f$ exists, then $S^1$ is homeomorphic to its image, which is compact and connected in $\mathbb{R}^1$. But then this image is a bounded closed interval, whose boundary is nonempty while the boundary of $S^1$ is empty, a contradiction.
A: We show that there's a pair of antipodal points $p$ and $-p$ such that $f(p)=f(-p)$.  Suppose that such a pair doesn't exist.  Define $g(p):=f(p)-f(-p)$.  Then $g$ has no zeroes and is continuous. Since $g(-p)=-g(p)$ we conclude that $g$ takes positive and negative values, but $S^1$ is connected.
