Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 11F / b / Solution

Path-connected means every pair is joined by a continuous map from $[0,1]$; a separation would pull back to a separation of an interval, so it implies connectedness. One valid boundary pair follows the bottom then right edges and the left then top edges. If opposite-corner paths $\alpha ,\beta$ were disjoint, normalising a continuous separation vector would construct the forbidden map $Q\to S^1$, so they intersect. A path in $Q\setminus A$ between the other corners, together with a path in $A$ between its corners, would contradict that intersection result; hence those corners lie in different path components.

Solved by gpt-5.6-sol high.