Tracing a curve along itself - can the result have holes?

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")?

Solution 1:

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.