Composition with function whose graph is everywhere dense in $\mathbb R ^2$ Let $f\colon \mathbb R \to \mathbb R$ be a function whose graph is everywhere dense in $\mathbb R ^2$.
Let $I\subseteq\mathbb R$ and $g\colon I\to \mathbb R$ be a surjective function.
I'm concluding since $g$ is surjective, the graph of $f \circ g$ is everywhere dense in $I\times \mathbb R$. Is this correct?
 A: seems plausible at first, but here is a counterexample: 
take I = $\mathbb{R}$, and define $g(x) = 0$ for $x < 0, g(x) = xsin(x)$ else.  Then $g$ is onto, but $f(g(x))$ is constant for $x < 0$.
A: As shown by the counterexample by user136920 surjectivity is not sufficient. But in case $g$ is also injective, i.e. bijective, and continuous the answer is yes.
Informal definiton: Denote by $G_f$ the graph of a function $f$.
Proof. Let $f \colon \mathbb R \to \mathbb R$ be a function whose graph is everywhere dense in $\mathbb R ^2$. Let $I\subseteq \mathbb R$ and $g \colon I \to \mathbb R$ be a continuous, bijective function. Let $\epsilon >0$ be given. Fix $(x,v) \in \mathbb R ^2$. $g$ is surjective so $u \in I$ exists, so that $g(u)=x$.
Since $g$ is continuous $\delta_1>0$ can be chosen so that $\lvert g(u_0)-g(u)\lvert<\frac{\epsilon}{\sqrt{2}}$ whenever $\lvert u_0 - u \rvert < \delta_1$.
Since $g$ is continuous and injective, a continuous inverse exists, which implies that a $\delta_2>0$ can be chosen, so that $\lvert u_0-u \rvert <\frac{\epsilon}{\sqrt{2}}$ whenever $\lvert g(u_0)-g(u) \rvert <\delta_2$.
Set $\eta = \min \{\delta_1,\delta_2\}$.
Because $f$ is everywhere dense in $\mathbb R ^2$, $x_0 \in \mathbb R$ can be chosen so that
$$ \lVert (x_0, f(x_0) - (x,v) \rVert = \lVert (x_0 - x, f(x_0) - v) \rVert < \eta $$
Then $\lvert x_0 - x \rvert < \eta$ and $\lvert f(x_0) - v \rvert < \eta < \tfrac{\epsilon}{\sqrt 2}$.
As mentioned earlier, because $g$ is surjective there exist $u,u_0 \in I$ so that $g(u)=x$ and $g(u_0)=x_0$. Remember $\lvert x_0 - x \rvert = \rvert g(u_0)-g(u) \lvert < \eta$ implies $\lvert u_0-u \rvert < \frac{\epsilon}{\sqrt 2}$.  It now follows that
$$ \lVert (u_0, f(g(u_0)) ) - (u,v)\rVert \leq \sqrt 2 \max \{ \lvert u_0 - u \rvert, \lvert f(g(u_0))-v \rvert \} < \epsilon  $$
This means for arbitrarily chosen point $p=(u,v)\in I \times \mathbb R$ and $\epsilon>0$,
$$B(p,\epsilon)\cap G_{f\circ g} \neq \emptyset$$
i.e. $p$ is a limit point of the graph of $f\circ g$; hence we conclude that every point in $I \times \mathbb R$ is a limit point of the graph. In other words the graph of $f\circ g$ is everywhere dense in $I \times \mathbb R$.
