Can any two points in the plane be assigned continuously a path between them that is transversal to two circles? More formally, I'm interested in continous functions
$$\Gamma : \mathbb{R}^2\times \mathbb{R}^2 \to ([0,1] \to \mathbb{R}^2)$$
such that $\Gamma(p,q)(0)=p$ and $\Gamma(p,q)(1)=q$, with the additional condition that for every $(p,q)\in \mathbb{R}^2\times \mathbb{R}^2$, we have that $\Gamma(p,q) \pitchfork X$, where $X$ is two circles of radius 1 centered in $(-2,0)$ and $(2,0)$. For the transversality condition to make sense, we require that $\Gamma(p,q)$ is smooth.
It can be seen that when $X$ is just one circle, this can be done, by composing the path that takes $p$ to the origin, and then goes back to $q$. Sadly, the same trick can't be applied to the two circles, at least as far as I can see. I suspect there is no function that satisfies these requirements, but I have not been able to show this.
I'm interested in this as it's a special case of topological complexity, and may give a strictly greater upper bound.
Edit: I'm assuming the compact open topology on $[0,1] \to \mathbb{R}^2$.
 A: It appears that no such map exists due to topological obstructions. Here's a rough sketch on one possible argument:
Let $C_-,C_+\subset\mathbb{R}^2$ be the left and right circles, and let $U=\mathbb{R}^2\setminus(C_+\cup C_-)$ be their compliment. Let $T\subset C^\infty([0,1],\mathbb{R}^2)$ denote the space of smooth paths in $\mathbb{R}^2$ transverse to $C_+\cup C_-$, with the compact open topology, and let $\widetilde{T}\subset T$ denote those whose endpoints do not lie in $C_+\cup C_-$. It suffices to show there is no continuous map $\Gamma:U\times U\to\widetilde{T}$ satisfying $\Gamma(p,q)(0)=p$, $\Gamma(p,q)(1)=q$. We can assume for the sake of contradiction that such a $\Gamma$ exists, and define some useful auxiliary functions:
Let $l:\widetilde{T}\to[0,1]$ be the "last intersection" function:
$$
l(\gamma)=\max\{t\in[0,1]:\gamma(t)\in C_+\cup C_-\}
$$
let $D_p:T\to\mathbb{R}$ be the miniumum Euclidean distance to $p\in\mathbb{R}^2$:
$$
D_p(\gamma)=\min\{d(p,\gamma(t)):t\in[0,1]\}
$$
and let $\theta_p:T_p\to\mathbb{R}$ be the winding number about $p\in\mathbb{R}^2$, where $T_p\subset T$ is the set of paths that do not pass through $p$:
$$
\theta_p(\gamma)=\int_0^1\frac{(\gamma_1(t)-p_1)\dot\gamma_2(t)-(\gamma_2(t)-p_2)\dot\gamma_1(t)}{\|\gamma(t)-p\|^2}
$$
Let $p_0=(-2,0)$ and $q_0=(0,0)$. Assume w.l.o.g. that the last intersection of $\Gamma(p_0,q_0)$ is with $C_-$.
Let $\alpha:[0,1]\to U\times U$ be a loop based at $(p_0,q_0)$ holding $p$ fixed and wrapping $q$ around $C_+$:
$$
\alpha(t)=\left(\binom{-2}{0},\binom{2-2\cos(2\pi t)}{-2\sin(2\pi t)}\right)
$$
Let $\varphi:[0,1]\to T$ be the rescaling of $\Gamma\circ\alpha$ to include only the segment between the last intersection and the endpoint. Note $\varphi(s)$ intersects the circles only at $\varphi(s)(0)\in C_-$
$$
\varphi(s)(t)=\Gamma(\alpha(s))((1-t)l(\Gamma(\alpha(s)))+t)
$$
Now, one needs to show the following:

*

*$D_p$, and $\theta_p$, $\varphi$, and $l\circ\Gamma$ are continuous on their domains.

*If $\varphi:[0,1]\to T$ is a continuous family of "good" curves and one $\varphi(s_0)$ lies strictly outside of $C_+$, then all $\varphi(t)$ lie strictly outside $C_+$ (equivalently for $C_-$) (this follows from continuity of $D_p$).

*$\theta_{(0,2)}\circ\varphi$ is well-defined and continuous.

*$\theta_{(0,2)}(\varphi(0))\neq\theta_{(0,2)}(\varphi(1))$, which is the desired contradiction.

