Revising for an exam - past paper question. I'm first asked to explain why a triangulation os $S^2$ must contain at least 4 vertices. I believe this might have to do with the fact that 3 vertices would give a triangle, and so fundamental group of this is that of the circle, i.e. additive group of integers. Can't explicitly reason as to why it can't be 2 or 1 currently though...
Then the question above is given - I can assume that $S^n$ can be triangulated and that $S^n$ minus a point is homeomorphic to $R^n$ - but not that X and Y can be triangulated.
So I triangulate $S^2$ and $S^3$ to get K and L resp., and also then using the homotopy equivalences from X, Y to $S^2$ and $S^3$ I can form a map |K|$\to$$S^2$$\to$X$\to$Y$\to$$S^3$$\to$|L|. By the simple approximation theorem, I can find a subdivision |K'| and a simplicial map $g$: K'$\to$L such that |$g$| is homotopic to this map.
I' not entirely clear where to go from here though. I have a feeling I'll have to remove a point and use the fact that $S^n$ minus a point is homeomorphic to $R^n$ but I'm unclear on how this actually works/whether its correct.
Thanks a lot.
For the second question, it is equivalent to show that a map $f:S^2\rightarrow S^3$ is homotopically equivalent to a constant. (Take the map $S^2\to X\to Y\to S^3$) This map is not surjective, up to homotopy see the second answer in the reference, so it takes its values in $S^3-$ point, and the hint tell you that this space is homeomorphic to a point.
For the first question consider the Euler characteristic, if you have less than 3 vertices, the number of edges is inferior to the number of vertices and the Euler number is positive. Contradiction.
Continuous map between spheres