I would just like to check if my way of proving the following statement is correct.
For $f:\mathbb S^1\to \mathbb R^2$ \ $P$ [here $P \in \mathbb R^2$] one defines the linking number as it follows:
$Um(f,P):=deg$ $(z\to \frac {f(z)-P}{||f(z)-P||})$
The claim is next:
For a fixed point $P \in \mathbb R^2$,
$f_0,f_1:\mathbb S^1\to \mathbb R^2$ \ $P$ are homotopic iff $Um(f_0,P)=Um(f_1,P)$.
My idea of proof is that $f_0,f_1$ are homotopic iff there exists a homotopy $f_t$ $0\leq t\leq 1$ between them iff $\frac {f_t(z)-P}{||f_t(z)-P||}$ is a homotopy between $\frac {f_0(z)-P}{||f_0(z)-P||}$ and $\frac {f_1(z)-P}{||f_1(z)-P||}$ iff $Um(f_0,P)=Um(f_1,P)$
My first question is if the last homotopy is indeed a homotopy and my second question is, if so, what can guarantee that $P \notin im(\frac {f_t(z)-P}{||f_t(z)-P||})$ for any such $t$?