TITLE: If the graph $G(f)$ of $f : [a, b] \rightarrow \mathbb{R}$ is path-connected, then $f$ is continuous.
QUESTION [8 upvotes]: Yesterday I woke up thinking about this question, and I believe I have a proof, but I'm not sure of its validity.
Let $\gamma : [a, b] \rightarrow \mathbb{R}^{2}$ be a path from $(a, f(a))$ to $(b, f(b))$. First of all, I will show we may take WLOG $\gamma$ to be injective. Consider a point $\alpha = (p, q) \in \mathbb{R}^{2}$. Consider the set $S = \{x \in [a, b]; \gamma(x) = \alpha\}$, which we suppose non-empty. Let $c = \inf S$, $c' = \sup S$. By the continuity of $\gamma$, we have $\gamma(c) = \gamma(c') = \alpha$. We may now consider the equivalence relation $\sim$ on $[a, b]$ which sets $x \sim x$ and $z \sim y$ for $z, y \in [c, c']$. The quotient space $[a, b] / \sim$ is homeomorphic to an interval (since $a < c \leq c' < b$, by the continuity of $\gamma$), and the induced map $\bar{\gamma} : [a, b] / \sim \rightarrow \mathbb{R}^{2}$ may be taken as a path which passes through $\alpha$ only once.
Since $[a, b]$ is compact and $\mathbb{R}^{2}$ is Hausdorff, $\gamma$ is a homeomorphism onto its image. Clearly $\gamma([a, b]) \subset G(f)$. We must now show this is an equality, which is done via the intermediate value theorem. Consider $\pi_{1} : G(f) \rightarrow \mathbb{R}$ to be the projection onto the first coordinate, which is continuous. Consider $\phi = \pi_{1} \circ \gamma : [a, b] \rightarrow [a, b]$, which also is continuous. Since $\phi(a) = a, \phi(b) = b$, $\phi([a, b]) = [a, b]$, and therefore $\gamma([a, b]) = G(f)$. Thus $f$ is a function whose graph is homeomorphic to its domain. We will show in the next paragraph that this implies the continuity of $f$.
Indeed, suppose $g : [a, b] \rightarrow \mathbb{R}$ has a homeomorphism $H : [a, b] \rightarrow G(g)$. This induces a map $T : [a, b] \rightarrow [a, b]$ such that $H(x) = (T(x), g(T(x))$. We note $T$ is continuous since $T = \pi_{1} \circ H$. It is clearly seen $T$ is bijective, and thus a homeomorphism since [a, b] is compact and Hausdorff. Since $g \circ T$ is continuous, $g \circ T \circ T^{-1} = g$ also is.

REPLY [4 votes]: Here is a very simple proof which reduces everything to a couple of standard general topology theorems. Let $G=G_f\subset {\mathbb R}^2$ denote the graph of $f: [a,b]\to {\mathbb R}$. Observe that the projection map $p: G\to [a,b]$ (given by the projection to the first component) is a continuous bijective map, its inverse is $x\mapsto (x, f(x))$. Hence, it suffices to show that $p$ is a homeomorphism. You probably already know that a continuous bijective map from a compact space to a Hausdorff space is a homeomorphism. Hence, it suffices to show that $G$ is compact. Since $G$ is path-connected, there exists a continuous map $\gamma: [0,1]\to G$ sending $0$ to $A=(a,f(a))$ and $1$ to $B=(b,f(b))$. I claim that this map is surjective. If not, then composing it with $p$ we obtain a continuous non-surjective map $[a,b]\to [a,b]$ which fixes the end-points, contradicting the Intermediate Value Theorem. Thus, $\gamma$ is surjective, which implies that $G$ is compact (a continuous image of a compact space is compact). qed 
Edit. This proof also suggests a higher-dimensional generalization of this result:
Lemma. Let $B^n$ be the $n$-dimensional closed unit ball, $f: B^n\to X$ a map to a Hausdorff topological space such that the restriction of $f$ to $\partial B^n$ is continuous. Then the following are equivalent:

$f$ is continuous. 
The graph $G_f$ of $f$ is contractible. (Path connected is, obviously, insufficient.) 
The restriction $(id, f): \partial B^n\to G_f$ defines a null-homologous spherical cycle $g: S^{n-1}\to G_f$ (more precisely, the image of the homotopy class $[g]$ under the Hurewicz homomorphism $\pi_n(G_f)\to H_n(G_f)$ is zero). 

The proof is essentially the same as written above: Assume (3). If $\sum_i a_i \sigma_i$ is a singular $n$-chain whose boundary is the spherical cycle $g$, then the union of images of the singular simplices $\sigma_i$ has to be the entire $G_f$ (otherwise, it misses a point $(q,f(q))$ and, hence, the boundary of $B^n$ is null-homologous in $B^n -\{q\}$, which is impossible). But then $G_f$ is compact, hence, the projection map $G_f\to B^n$ is a homeomorphism, hence, $f$ is continuous.  The implications (1)$\Rightarrow$(2) and  (2)$\Rightarrow$(3) are clear.