Take a unit square in $\mathbb{R}^2$ and let $\gamma \colon [0,1] \to [0,1]^2$ be a continuous path from $[a,0]$ to $[0,b]$ where $0\le a,b < 1$ are constants. So we can think of $\gamma$ as cutting the square into two pieces as per the graph below.
Now let $f \colon [0,1]^2\to[0,1]^2$ be a continuous function with the property that $f(\gamma(t)) = 0$ for all $t\in[0,1]$. Let $\le$ denote the element-wise partial order on $\mathbb{R}^2,$ and note that $f(\gamma(t)) \le \gamma(t)$ for all $t \in [0,1].$ I want to prove that $f$ has a fixed point when its domain is restricted to the set $$\Gamma = \{(x,y) \in \mathbb{R}^2 \colon (x,y) \le \gamma(t) \text{ for some } t \in [0,1]\}.$$ If $f(\Gamma) \subseteq \Gamma,$ there is likely some standard fixed point theorem we can apply (perhaps Lefschetz?). But if $f(\Gamma) \not\subseteq \Gamma$ then we run into problems.
Intuitively, if we have that $f(x,y)$ is outside of $\Gamma$ for some $(x,y)$ then it must be that $f(x,y) < \gamma(t)$ for some $t$. But since $f(\gamma(t))> \gamma(t),$ as we move away from the line traced out by $\gamma$ and towards the origin, there must be some point where the inequality "swaps", i.e. a fixed point!
Is there a fixed point theorem I can use here to prove this? Or perhaps the intuition is misleading and there is not necessarily a fixed point (or we need more structure on the problem)?
This is the simplest version I could write down of a more general problem which can be posed in $\mathbb{R}^n,$ so any solution which offers a natural generalization to $n$ dimensions is preferred. Thanks!