Degree of an antipodal map

Let $f:S^n\to S^n$ a continuous map, $n>0$; we consider the induced homomorphism $f_* : H_n(S^n)\to H_n(S^n)$, and, recalling $H_n(S^n)\simeq\mathbb Z$, define $deg(f)\doteq f_*(1)$.

I'm asked to find the degree of the reflection map $f:(x_0,\dots,x_n)\mapsto (x_0,\dots,x_{n-1},-x_n)$; my main problem is that I should do it only using tools from singular homology theory: I have not dealt with the concept of orientation of a manifold, nor I know anything about reflections on an hyperplane (I've browsed previous questions, and these seems to be the suggested tools to address the quandary).

It looks like the only possible way to proceed is to track the way singular simplices $\sigma:\Delta_n\to S^n$ are influenced by the map $f$, and then try to see how their homology class changes: but it's quite an impervious path, and I don't see how to proceed.

Any hints?

Edit: I'm also having problem dealing with this (apparent) contradiction: consider the maps $f,\pi:S^1\to S^1$ with $f(x_1,x_2)=(-x_1,x_2)$ and $\pi(x_1,x_2) = (x_2,x_1)$; then $f\cdot \pi\cdot f\cdot\pi = -id$, thus apparently $deg(f\cdot\pi))^2=-1$...

• The map you wrote down is a reflection, not the antipodal map. By way of contrast, the antipodal map would send $(x_0,...,x_n)$ to $(-x_0, ..., -x_n)$ (every coordinate is negated). If you want to go the simplices method, I suggest finding a nice triangulation so that your desired map sends simplices to simplices. So, in the case of $S^1$, I'm thinking of a diamond shape, for $S^2$ I'm thinking of two triangular based pyramids (sans base), stuck base to base. Try to generalize this. – Jason DeVito Jan 16 '15 at 14:12
• I don't think your definition of the antipodal map is right. Usually it refers to the map $x \mapsto -x$. $(x_0, \ldots, x_n) \mapsto (x_0, \ldots, x_{n-1}, -x_n)$ is just a reflection in the last variable. Anyway, Hatcher computes the degree on p. 134 here. – Viktor Vaughn Jan 16 '15 at 14:12
• Thank you both, I've edited the question: it was just a reflection. I'll try to think of possible triangulations - I've had a look at Hatcher's approach, but he uses simplicial homology theory, which I'm not familiar with... – nelv Jan 16 '15 at 14:23
• As per your edit, on $S^1$, the map $-id$ has degree $1$, not $-1$. – Jason DeVito Jan 16 '15 at 14:27

Realize ${\mathbb S}^n$ as ${\mathbb D}^n / \partial{\mathbb D}^n$ with basepoint $\overline{\partial{\mathbb D}^n}$, you have a pinch map ${\mathbb S}^n \to {\mathbb S}^n\vee{\mathbb S^n}$ collapsing $\{x_n = \frac{1}{2}\}$ to a point, and which you can prove to induce the diagonal ${\mathbb Z}\to{\mathbb Z}\oplus{\mathbb Z}$ on $n$-th singular homology. Moreover, for any two pointed maps $f,g: {\mathbb S}^n\to{\mathbb S}^n$, the map $f\vee g: {\mathbb S}^n\vee{\mathbb S}^n\to{\mathbb S}^n$ induces the map $(\text{deg}(f),\text{deg}(g)): {\mathbb Z}\oplus {\mathbb Z}\to {\mathbb Z}$ on $n$-th homology. Putting both together, you get that the map $f+g:{\mathbb S}^n\to{\mathbb S}^n$ defined as the composition $$f+g:={\mathbb S}^n\to{\mathbb S}^n\vee{\mathbb S}^n\xrightarrow{f\vee g}{\mathbb S}^n$$ has degree $\text{deg}(f)+\text{deg}(g)$. Now, taking $f=\text{id}$ and $g$ the map flipping the $n$-th coordinate, you can check that $f+g$ is nullhomotopic, so ...?
• What is $f+g$ exactly? – Matt Samuel Jan 16 '15 at 14:32
• @MattSamuel: In general, you mean? On $\{x_n \leq \tfrac{1}{2}\}$, it maps $[x_1,...,x_n]$ to $f([x_1,...,x_{n-1},2x_n])$, while on $\{x_n \geq\tfrac{1}{2}\}$ it maps it to $g([x_1,...,x_{n-1},2x_n-1])$. – Hanno Jan 16 '15 at 14:36
• I hoped that one could understand the answer both intuitively and formally without knowing anything about homotopy groups, because the definition of $f+g$ is simple to write down as the composition of the pinch map and $f\vee g$ – Hanno Jan 16 '15 at 14:42
• What was unclear was the notation $f+g$. I didn't see an explicit definition of it. What I see is "putting both together" then $f+g$. – Matt Samuel Jan 16 '15 at 14:57