Mapping a torus onto a sphere Suppose a torus $T$ (with the usual topology) decomposed into 28 hexagonal 2-cells like so:

If I fold each hexagon into a triangle like so:

and stack them in groups of 7, I can construct a surjection $f: T \rightarrow f(T)$ which maps $T$ onto a tetrahedron $f(T)$ preserving the incidence structure of the 2-cell decomposition, i.e. if a vertex $v$ is incident to an edge $e \in T$, then $f(v)$ will be incident to $f(e)$.
Now, the tetrahedron being homeomorphic to the sphere, I know $T$ cannot possibly be a covering space for $f(T)$. However, I am having difficulty identifying where it fails to meet the requirements. It appears to me like: any given point $x \in f(T)$ will have a neighborhood $U \subseteq f(T)$ such that $f^{-1}(U)$ is the union of exactly 14 open sets in $T$, which will be a local homeomorphism for each set as the torus and sphere are both closed surfaces, so we can always find a $U$ homeomorphic to the open unit disc. What am I missing?
 A: A continuous map $f : M \to N$ between compact connected manifolds is a covering map if and only if it is a local homeomorphism, meaning that each point $x \in M$ has a neighborhood $U \subset M$ such that $f \mid U$ is a homeomorphism onto its image. So one way to disprove that $f$ is a covering map is to identify a point $x \in M$ at which $f$ fails to be locally injective.
Now, your description of the map is too vague to pinpoint exactly where one may find a point at which $f$ is not locally injective. However, some point near the center of that hexagon you have drawn seems like a good candidate. The boundary of the hexagon wraps 2 times around the boundary of the triangle. Depending on the exact details of that hexagon-to-triangle map, perhaps the center point of the hexagon is where local injectivity fails: one can imagine that concentric hexagons, shrinking down to the hexagon's center, are each wrapping 2 times around concentric triangles, shrinking down to the triangle's center.
The hexagon-to-triangle map would, in this case, be called a branched covering map, with degree 2 branching over the center of the hexagon.
And there are, indeed, many ways to construct branched covering maps of the torus over the sphere.
