homotopy can't extend to one point compactification. Let $f,g: \mathbb{C} \rightarrow \mathbb{C}$ be the functions $f(z) = z^n$ and $g(z) = z^m$, $n\neq m$.  Taking the one point compactification both of these functions extend to continuous functions $ f^1, g^1: \mathbb{C}^1\rightarrow \mathbb{C}^1$ where $\mathbb{C}^1 = \mathbb{C} \cup \{\infty\}$ and $f^1(\infty) = g^1(\infty) = \infty$.
In the complex plane, there is a linear homotopy $F:\mathbb{C}\times I \rightarrow \mathbb{C}$ from $f$ to $g$ given by $F(z,t) = (1-t)f(z) + tg(z)$. We can then define a map $G:\mathbb{C}^1\times I\rightarrow \mathbb{C}^1$ by $G(z,t) = F(z,t)$ and $G(\infty,t) = \infty$. I want to know why this cannot be a continuous map using basic point-set topology.
My thoughts: The map's restriction to $\mathbb{C}$ is clearly continuous, so something fails on the set $\infty \times I$. Let $K_N = \{z\in \mathbb{C}| \,|z|\leq N \}$ then $U_n = K_n^c$ is an open neighborhood of $\infty$, and I think these sets are a neighborhood base so the function should fail to be continuous on one of these.
(I can't seem to write \hat in latex, anyone know how to fix this?)
 A: You're asking to disprove the claim that
$$ \lim_{z \to \infty, t \to t_0} F(z, t) = \infty$$
The usual method is to approach the point from multiple paths. Let's boldly assume there is a path where the limit would come out to zero. Let's be even more bold and say the function is identically zero along the path.
Along such a path, we have $t = 1 / (1 - z^{m-n})$ (or, if $z^{n-m} \neq 1$ and some conditino I haven't worked out).
It's easy to find such a path now: just choose $w$ so that $w^{m-n}$ is a negative real number, and let $f(x) = w^{m-n} x$ where $x$ is a real number, and let $x \to \infty$.
$$\lim_{x \to \infty} F(w x, 1 / (1 - (wx)^{m-n})) = 0 $$
which is not the value of $G(\infty, 0)$.
A: Using algebraic topology, it is easy to prove that there is no homotopy between $f$ and $g$ by comparing the degree of the induced homomorphisms between the homology groups of $\hat {\mathbb C}$. If you want to get to the core of why the homotopy $F$ you defined cannot be extended, then point-set-topology will do it.
Assume that $G:(\mathbb C\cup\{\infty\})\times I\to\mathbb C\cup\{\infty\}$ is a homotopy that fixes $\infty$ and extends $F$. By the universal property of the quotient space $Q$ which identifies $\{\infty\}\times I$ to a point, it induces a continuous $\hat G:Q=(\mathbb C\times I)\cup\{\infty\}\to\mathbb C\cup\{\infty\}$, where the domain actually carries the compactification topology which coincides with the quotient topology. Now a map between locally compact Hausdorff spaces $f$ can be extended to a map $\hat f$ between their one-point-compactifications if and only if $f$ is proper, i.e. the preimage of a compact set is compact.
We'll see that $\hat G|_{\mathbb C\times I}=F$ is not proper, because the preimage of $\{0\}$ is not bounded:
Let $n>m>0$ in $\mathbb N$. There is a $\phi$ such that $(n-m)\phi=\pi$. Define $z=re^{i\phi},\  r>0$. We have $z^n/z^m=r^{n-m}e^{i\phi}=-r^{n-m}$, so $z^n=-r^{n-m} z^m$. You want $tz^m+(1-t)z^n=0$ which implies $t/(1-t)=r^{n-m}$. So no matter how big $r$ is, there is always a $t\in(0,1)$ satisfying this equation.
