How do you know at least one face is not simply connected on a polyhedra?

Question

if it has 14 vertices, 21 edges and 9 faces, its boundary is a single surface and there is at least one hole. I dont understand.

Answer 1

If all the faces are simply connected, then the vertices, edges, and faces (the boundary complex of the polyhedron) form a CW-complex. The Euler characteristic is $9 - 21 + 14 = 2$, so this complex is topologically a sphere. But this contradicts the assumption that there is a hole through the polyhedron.

Therefore, at least one face is not simply connected.