I'm trying to get acquainted with homotopy theory. In particular I'm now reading Chapter 3 of Dubrovin, Fomenko, Novikov. Modern geometry—methods and applications: Part II, Springer, 2012.
I have a perhaps silly doubt on the notions of proper homotopy and topological degree of a function (here denoted by $\deg$). Specifically, my doubt concerns the following very simple example.
Consider the real-valued polynomials $p(x):=x^2$ and $q(x):=x^3$ and consider the map $$ f(x,a):= ap(x) + (1-a)q(x),\quad a\in[0,1]. $$ Clearly, $p(x)$ and $q(x)$ are proper maps from $\mathbb{R}$ to $\mathbb{R}$. Moreover, it holds $\deg(p)=0$ and $\deg(q)=1$ and (I would say that) $f(x,a)$ is a continuous proper homotopy between $p(x)$ and $q(x)$ (I think that here I'm mistaken but I don't see what the error is). A well-known theorem states that properly homotopic proper maps have the same degree, therefore we have $\deg(p)=\deg(q)$. But this contradicts the fact that $\deg(p)=0$ and $\deg(q)=1$!
Could you please explain to me where my mistake is?
Thanks in advance.
The issue is that your proposed homotopy is not proper!
Consider, for example, the preimage of $0$.
$ax^3 + (1-a)x^2 = 0$ has the nontrivial solution $x= \frac{a-1}{a}$, for $a \not= 0$, and gives $$f^{-1}(0) = ([0] \times [0,a]) \bigcup \{(\frac{a-1}{a}, a) \ | \ a \in (0, 1] \}$$
This preimage isn't compact since $\frac{a-1}{a}$ blows up to infinity for $a$ small.