Theorem 9.5 from Introduction to Topology by Colin Adams and Robert Franzosa reads,
Theorem 9.5. For each circle function, $f:S^1 \to S^1$, there exists a unique $n \in \mathbb{Z}$ such that $f$ is homotopic to $c_n(\theta) = n\theta$.
My Question: How can I calculate the degree of a circle function, $f:S^1 \to S^1$ defined by $f(\theta) = c$, where $c$ is a constant?
I tried to proceed in the following way. I defined a homotopy $F: S^1 \times I \to S^1, F(\theta, t) = (1-t) \, c + tn\theta, \, \theta \in S^1, \, t \in I = [0,1]$ between the circle functions $f$ and $c_n$ such that $F(\theta, 0) = f(\theta) = c$ and $F(\theta, 1) = n\theta$.
Best Answer
Your homotopy fails because for $t\in (0,1),$ $F(\theta,t)$ fails to map to $0$ and $2\pi$ to the same element of the circle (for instance $e^{i0}\neq e^{i\frac{1}{2}2\pi}$ when $t=\frac{1}{2}$). Thus, it fails to descend from $I$ to the quotient $S^1$, even though your formula looks suggestive (you need to be very careful with just treating $S^1$ as $I$ when building homotopies since any two maps from $I$ into $S^1$ are homotopic).
Now, hopefully, your argument works for some $n$ and, indeed, my above objection doesn't hold for $n=0,$ because the map $\theta\mapsto (1-t)c$ clearly maps $0$ and $2\pi$ to the same number. Hence, in this case, your proposed homotopy is, indeed, a homotopy, and you get that the degree of a constant map is $0$.