Let $a\colon [0,1] \to D^2 = \{z \in \mathbb{C}\colon |z| \leq 1 \}$ be a continuous map with $|a(0)| = |a(1)|$ and further assume that $a$ is an embedded - so $a$ is an embedded arc. Let $f\colon D \to D$ given by $f(z) = z^2$.
Under what conditions is $f \circ a$ still injective? It seems to me that a necessary and sufficient condition is that $a$ only intersect at most one of each pair radial arcs between $0$ and the unit circle that are identified by $f$, but it is unclear to me how to prove this.
Consider $f \circ a \colon [0, 1] \to D^{2}.$ Then, $f \circ a$ is not injective if and only if there exist distinct $x, y \in [0, 1]$ such that $$(f \circ a)(x) = (f \circ a)(y).$$ Note that $$(f \circ a)(x) = f(a(x)) = a(x)^{2},$$ and similarly $$(f \circ a)(y) = a(y)^{2}.$$ So, we have $$a(x)^{2} = a(y)^{2}.$$ Since $x \neq y$ and $a$ is injective, we have $a(x) \neq a(y).$ So, it follows that $$a(x) = -a(y).$$
So, $f \circ a$ is not injective if and only if there exist distinct $x, y \in [0, 1]$ such that $a(x) = -a(y).$ Therefore, $f \circ a$ is injective if and only if there do not exist distinct $x, y \in [0, 1]$ such that $a(x) = -a(y).$