Let $\varphi:[0,1]\to\Bbb R^2$ be a continuous curve (not necessarily injective) with $\varphi(0)=(0,0)$. Let $f:[0,1]^2\to\Bbb R^2$ be defined as $f(s,t)=\varphi(s)-\varphi(t)$.
Question: Is the image $f([0,1]^2)$ always simply connected?
The set $f([0,1]^2)$ can be thought of as the trace of the mirror curve $-\varphi$ when slided along $\varphi$. See the image below for an example.
$\qquad\quad$
This problem came up to me in some non-trivial topological context but I consider it interesting in its own right. I am optimistic it can be proven with sufficiently advanced topological machinery, but might there be an elementary proof (as elementary as the term "simply connected")?
There can be holes, even for very reasonable curves. For example, a $\varphi$ like
will produce $f([0,1]^2)$ like
which is clearly not simply connected.