When do we have Brouwer fixed points only on the corner?

Question

Suppose $f:[0,1]^n\rightarrow[0,1]^n$ is a continuous function. When does $f$ only have integral fixed points where each coordinate of the fixed point is either 0 or 1. In one-dimensional case, it means that the curve of $f$ never crosses the $45^\circ$ line, then we only have $f(0)=0$ or $f(1)=1$ as fixed points. Are there any results on this problem?

Answer 1

The characterization you give in 1 dimensions is a restatement of $\text{fixed points}(f) = \{ x : (x,y) \in \Delta \cap \Gamma_f \} \subseteq \text{Corners}$, where $\Delta$ is the diagonal (graph of identity function), and $\Gamma_f$ is the graph of $f$.

Here is why this is a restatement of your characterization of the $1$ dimensional case: $\Delta$ is the 45 degree slope line, and by 'never crossing' I think you mean that the graph of $f$ intersects $\Delta$ only at its endpoints, which are the points of $\Delta$ with $x$ coordinates given by the corners.

The equation $\text{fixed points}(f) = \{ x : (x,y) \in \Delta \cap \Gamma_f \subseteq X \times X\} $ is true for any function $f : X \to X$ - it follows formally by unwrapping the definitions.

So, you can generalize your characterization to the containment $\text{fixed points}(f) = \{ x : (x,y) \in \Delta \cap \Gamma_f \} \subseteq \text{Corners}$. This is equivalent to stating that $\Gamma_f$ intersects $\Delta$ only at the vertices of $\Delta$, since $\Delta$ is isomorphic to $[0,1]^n$ (as polytopes) under the map that projects to the $x$-coordinate.