Finding the degree of a map

Question

I am having trouble computing the degree of a certain map using the fact that $f: N \rightarrow M$ where $M$ and $N$ are both $n$-dimensional manifolds induces a homomorphism between the nth cohomology groups. Also if there is a more algebraic topology approach as opposed to differential manifolds approach I would be interested in that as well.

The map is.

Consider the map $f : R^2 \rightarrow R^2$, $f(x, y) = (x^2 - y^2, 2xy).$ Find deg$(f)�.$

Thank you in advance.

Answer 1

You cannot use the cohomological approach to compute the degree of your map $f\colon \mathbb R^2 \to \mathbb R^2 $ because $\mathbb R^2$ is a non-compact manifold and $H^2(\mathbb R^2, \mathbb Z)=0$.
However your map restricts to a $2$-sheeted covering map $f_0\colon \mathbb R^2\setminus \{0\} \to \mathbb R^2 \setminus \{0\}$, so that the only reasonable degree you can attribute to $f$ is $2$.

A more striking way to compute the degree is to notice that after identification of $\mathbb R^2$ with $\mathbb C$, your map is nothing else than the squaring map $f\colon \mathbb C \to \mathbb C:z\mapsto z^2 $.
Complex analysts then say that you have a ramified covering of order $2$.

Another approach is to notice that $f$ can be extended to $\mathbb P^1(\mathbb C)$, the Riemann sphere, as a holomorphic map $F\colon \mathbb P^1(\mathbb C)\to \mathbb P^1(\mathbb C)$ by decreeing that $F(\infty)=\infty$.
The map $F$ is then a non-constant morphism of Riemann surfaces.
As such it has a degree which is (surprise,surprise!) $2$.

Yet another variant would be the Lefschetz approach, sketched by Arthur in his comment.

A personal opinion
I like your problem because you have to interpret the question in order to find an answer.
This is closer to real, research mathematics than more scholastic exercises which have a standard, uniquely defined but sometimes boring solution.

Answer 2

I like the answer by Georges. I would like to give an elementary answer. $\mathbb{R}^2$ is not compact, but $f$ is a proper map. Since $(0,0)$ is the only singular point and $\mathbb{R}^2\setminus\{(0,0)\}$ is connected, then the degree is constant. We choose a particular regular point $(1,0)$. Since $f^{-1}(1,0)=\{(-1,0),(1,0)\}$ (that one can deduce also from the fact that $f$ is the map $z\mapsto z^2$). In the other hand, $\mathrm{sign}\det D_xf>0$, for all $x\in\mathbb{R}^2\setminus\{(0,0)\}$, then $\mathrm{deg}\,f=2$.