I am trying to understand lens spaces. While understanding the basic definition is rather easy, everything that follows appears to be quite hard. I wanted to understand the classification up to homotopy equivalence, which is outlined as two exercises in Hatcher's book on algebraic topology, page 310 in Chapter 3.E and page 391 in Chapter 4.2.
There I am given a very specfic map between spheres $f\colon S^{2n-1}\to S^{2n-1}$ defined by $$f(r_1 e^{i\theta_1},\dots,r_n e^{i\theta_n})=(r_1 e^{ik_1\theta_1},\dots,r_n e^{ik_n\theta_n}).$$ For this map I need to compute the mapping degree, which is equal to the product $k_1 \cdot\dots \cdot k_n$. For these computations I have never actually seen a general way to do them. I know the rules for mapping degrees, but do not see how they apply, since I have only ever proven things about degrees, but never computed them. How do I tackle such a problem? How do I even begin?
Cheers
I think I found a nice answer to my question which doesn't involve any calculus whatsoever, but I would like confirmation that this actually works. The argument goes as follows:
The only degree computation I have ever actually seen was the degree of the map $S^1\to S^1,\,z\mapsto z^k$ with degree $k$. Now looking at the original map, I can guarantee, by permuting the entries, that above $f$ can be written as a composition of functions which only act on one entry of $S^{2n-1}$. The degree of $f$ can then be calculated as the product of the degrees of the single entry action. Thus I only need to consider the case $k_1=k$ and $k_i=1$ for $i\neq 1$. So far I am very certain that my argument works. The next step though might be a problem:
Isn't the above map with only one $k_i\neq 1$ exactly the $2n-1$-fold suspension of the exponentiation on $S^1$? Can anyone verify this claim?