Question 2 on page 11 of J. P. May's "A Concise Course in Algebraic Topology" reads:
Show that any map $f:S^1 \rightarrow S^1$ with $\text{deg}(f)\neq 1$ has a fixed point.
I have shown that if $f$ has no fixed point, then $\text{deg}(f)=-1.$ This contradicts the desired result. Where is my mistake? To be clear, I am $\textit{not}$ looking for a solution to the problem.
MY ATTEMPT: Suppose $f$ has no fixed point. Define the homotopy $f_t:S^1 \rightarrow S^1$ by \begin{align*} f_t(x)=\frac{(1-t)f(x)-tx}{|(1-t)f(x)-tx|}. \end{align*} This is a well-defined homotopy from $f(x)$ to $-x$ so long as the denominator does not vanish.
To see this is the case, suppose for contradiction we have $t\in I$ and $x\in S^1$ such that $(1-t)f(x)-tx=0.$ This implies $|(1-t)f(x)|=|tx|,$ which implies $|1-t|=|t|,$ which implies $t=1/2.$ Plugging this into $(1-t)f(x)-tx=0,$ we obtain $f(x)/2=x/2.$ This contradicts our assumption that $f$ has no fixed point.
We conclude that $f_t$ is indeed a well-defined homotopy from $f(x)$ to $-x,$ showing that $\text{deg}(f)=-1.$
You showed correctly that $f$ is homotopic to the map $-id : S^1 \to S^1, -id(x) = -x$. But it is not true that $\deg (-id) = -1$. In fact, $-id$ is a rotation (angle of rotation = $\pi$) which is homotopic to $id$. An explicit homotopy from $id$ to $-id$ is given by $$H(x,t) = e^{\pi i t}x.$$