Definition: A map $f:X\to Y$ between two topological spaces $X$ and $Y$ is said to be proper if $f^{-1}(K)\subseteq_\text{compact} X$ for every $K\subseteq_\text{compact} Y$. A homotopy $H:X\times [0,1]\to Y$ is said to be proper homotopy if $H$ is a proper map.
Problem: Consider the plane $\Bbb R^2$ and two maps $\text{Id},\overline{\text{Id}}:\Bbb R^2\to \Bbb R^2$ defined as
follows $\text{Id}(x,y)=(x,y)$ and $\overline{\text{Id}}(x,y)=(x,-y)$
for all $(x,y)\in \Bbb R^2$. Does there exist a proper homotopy
$H:\Bbb R^2\times [0,1]\to \Bbb R^2$ such that $H(-,0)=\text{Id}$ and
$H(-,1)=\overline{\text{Id}}$?
Note that any two maps of $\Bbb R^2=\Bbb C$ are homotopic via a convex combination. For example $G:\Bbb R^2\times [0,1]\to \Bbb R^2$ defined by $G\big((x,y),t\big)=\big(x,(1-2t)y\big)$ is a homotopy between $\text{Id}$ and $\overline{\text{Id}}$, but $G$ is not a proper homotopy.
Note also that, $\overline{\text{Id}}\circ \overline{\text{Id}}=\text{Id}$. Now, if we use compactly-supported cohomology then $H^2_\text{c}(\Bbb R^2;\Bbb Z)=\Bbb Z$ and we can talk about degree of a proper map: $\deg_\text{c}(\text{Id})=1$ and $\deg_\text{c}(\overline{\text{Id}})\cdot\deg_\text{c}(\overline{\text{Id}})=1$ i.e. $\deg_\text{c}(\overline{\text{Id}})=\pm 1$. Note that $\deg_\text{c}$ is invariant under proper homotopy. My guess is that $\deg_\text{c}(\overline{\text{Id}})=-1$, but I have no proof. Is there any other approach?
Thank you.
Best Answer
Suppose we have a proper homotopy $f_t:\mathbb C\times I\rightarrow \mathbb C$ with $f_0(z)=z$ and $f_1(z)=\bar z$.
We can extend this to a homotopy of the Riemann sphere $\mathbb C\mathbb P^1$ by defining $f_t(\infty)=\infty$ for all $t\in I$. Then continuity of $f_t:\mathbb C\mathbb P^1\times I\rightarrow \mathbb C\mathbb P^1$ follows from the properness of our original homotopy. But now $f_0$ is the identity and $f_1$ is a reflection of the Riemann sphere, so they have degrees $1$ and $-1$, respectively. This is a contradiction because degree is a homotopy invariant.