Path connectedness of graph and domain implies continuity? I'm trying to produce an example to show that a path-connected graph $G_f$ of a function $f$ whose domain is also path-connected does not imply the function is continuous. 
Does this example work:
Consider The function $f:[\theta,2\pi] \rightarrow \mathbb{R}^2$ where $f$ is defined in the following way: 
$f(\theta) = (cos(\theta),sin(\theta))$ for $0 < \theta < 2\pi,$ 
$f(0) = (0,1),$ 
and $f(2\pi) = (0,-1).$ 
The graph produced is a circle without the point $(0,1).$ Moreover, it is has a jump discontinuity at $\theta = 0$ and $\theta = 2\pi.$ However, the domain $[0,2\pi]$ and the graph are path connected.
Is there an error somewhere? I doubt myself...
 A: Your example does not work because, as said in comments you are confusing the graph of $f$ with it's image. The graph of a function $f:X\to Y$ is the subset of $X\times Y$ defined by $$G_f=\{(x,f(x)), x\in X\}$$ and in your example it is not connected.
An easy example of a disctontinuous function with path connected graph, and defined on  a path connected set, is the logarithm
Set $X=\mathbb C\setminus\{0\}$ and $f:X\to \mathbb C$ given by a determination of the logarithm, namely for any complex number $z$ you choose the polar coordinates with $\theta\in[0,2\pi)$ (i.e. set $z=\rho e^{i\theta}$) and define
$$f(z)=f(\rho e^{i\theta})=\log(\rho)+i\theta.$$ 
This function is not continuous on positive reals numbers (where $\theta=0$). However, its graph is path connected because because for any two points $z,w\in X$ you can connnect them with an arc not passing trhough $\mathbb R^+$ (except maybe at endpoints) and this produce a path in the graph connecting $(z,f(z))$ to $(w,f(w))$.
A: Let $C$ be the unit circle, which is path connected. Define $f:C \to \mathbb R$ by setting $f(e^{it})=t,0\le t <2\pi.$ Then $f$ is discontinuous at $(1,0).$ However the graph of $f$ is the spiral $\{(e^{it},t):0\le t <2\pi\},$ which is path connected.  
