Transfer homomorphism in Algebraic topology

Question

I am studying Hatcher's Algebraic topology. I am reading 3G about Transfer homomorphism. But most of the results are deemed obvious and I don't understnad why.

"Let $\pi:\tilde X\to X$ be an $n$-sheeted covering space for some finite $n$. Then there are $n$ distinct lifts $\tilde \sigma : \Delta^n\to \tilde X$ of a map $\sigma: \Delta^n\to X$."

I don't understand this part. So if $n\geq 3$ then the $n$-simplex is just the $S^{n-1}$, which has trivial fundamental group, and hence A lift exists, that's all I got. If $n=2$, then the fundamental group is $\mathbb{Z}$, so I don't even know why a lift exists. Can you explain this? I omit some parts of the books so I think I miss the details somewhere.

Also, there are more parts I don't understnad, can you recommend me a book about the details for this part of Hatcher's book?

Answer 1

All simplices are contractible, as was pointed out, so lifts exist.

Now given a based covering $(\tilde X,x)\to (X,b)$ and a based space $(Y,y)$ together with a map $(Y,y)\to (X,b)$ (the notation means it's a based map) satisfying the requirements for a lift to exist, then a based lift is unique.

So if the covering is $n$-sheeted, there are as many unbased lifts as there are possible choices of basepoints $x\in \pi^{-1}(b)$, that is, $n$ : there is a unique based lift for each such $x$.

Answer 2

hmmmm if I remember well the simplex is not what you claim. a 0-simplex is a point, a 1-simplex is a line and so on. In particular the 3-simplex is the ball in $\mathbb R^3$ not $S^2$.

All these simplices have $\pi_1=1$ because they are contractible spaces.

There is this book which contains a lot of specific examples "Topology, geometry, and gauge fields" form Gregory L. Naber".