If $f,g: S^1\to S^1$ are homotopic maps, then $\deg f = \deg g$ Problem:

If $f,g: S^1\to S^1$ are homotopic maps, then prove that $\deg f = \deg g$. Does the converse hold?


First, I shall explain the definition of degree I am familiar with:

Let $f:S^1\to S^1$ be a continuous map with $f' = f\circ p$. Choose a lifting $\widetilde{f'}: I\to \mathbb R$ with $p \circ \widetilde{f'} = f'$. We define $$\deg f = \widetilde{f'}(1) - \widetilde{f'}(0)$$
It can be shown that $\deg f$ is independent of the choice of the lifting $\widetilde{f'}$ of $f$.

I have shown that $\deg f$ is independent of the choice of the lifting $\widetilde{f'}$ of $f$.

My effort: I have the following commutative diagrams in mind, where $p: t\mapsto e^{2\pi i t}$ is a covering map from $\mathbb R\to S^1$, $\mathbf{1} = (1,0) \in S^1$, and $f\stackrel{F}{\simeq} g$:

Define $F': I\times I \to S^1$ using $F: S^1\times I \to S^1$ as follows: $$F'(s,t) = F(p(s), t)$$
Then $F'$ is a homotopy between $f' = f\circ p$ and $g' = g\circ p$. Let us lift this homotopy to $\widetilde{F'}: I\times I\to \mathbb R$ using the covering map $p: \mathbb R\to S^1$, so that $p\circ \widetilde{F'} = F'$ is satisfied. $\widetilde{F'}$ is a homotopy between $\widetilde{f'}$ and $\widetilde{g'}$, liftings of $f'$ and $g'$ respectively. Since the degree of $f,g$ does not depend on the choice of liftings $\widetilde{f'}, \widetilde{g'}$ - I decided to choose arbitrary liftings. We know that $\deg f = \widetilde{f'}(1) - \widetilde{f'}(0)$ and $\deg g = \widetilde{g'}(1) - \widetilde{g'}(0)$, so we want to show $\widetilde{f'}(1) - \widetilde{f'}(0) = \widetilde{g'}(1) - \widetilde{g'}(0)$ in order to establish the claim.

Reference:

Unique Path/Homotopy Lifting Property: Let $p: E\to B$ be a covering map. Let $f: I\to B$ be a path, with $f(0) = b$ and $p(\tilde e) = b$. Then there is a unique lifting $\tilde f: I\to E$ with $\tilde f(0) = \tilde e$. Moreover, let $F:I\times I\to B$ be a homotopy with $F(0,0) = b$ and $p(\tilde e) = b$. Then there is a unique lifting $\tilde F: I\times I \to E$ with $\tilde F(0,0) = \tilde e$.

 A: Suppose $f,g:S^1\to S^1$ are homotopic. Let $F:S^1\times[0,1]\to S^1$ be a homotopy from $f$ to $g$. Define a map
$$\bar{F}:[0,1]\times[0,1]\to S^1$$
given by $(s,t)\mapsto F(q(s),t)$ where $q:[0,1]\to S^1$ is a quotient map.
Then there is a lift
$$\tilde{F}:[0,1]\times[0,1]\to\Bbb R$$
with $\tilde{F}(0,0) =0$ of $F$. Then $f$ and $g$ lifted to $\tilde{f}$ and $\tilde{g}$ with $\tilde{f}(0) = 0$ and $\tilde{g}(0)=:k$.
By the uniqueness of path lifting, $\tilde{F}(s,0) = \tilde{f}(s)$. Define
$$\bar{\gamma}_0:\{0\}\times[0,1]\to S^1$$
by $(0,t)\mapsto \bar{F}(0,t)$ and
$$\bar{\gamma}_1:\{1\}\times[0,1]\to S^1$$
by $(1,t)\mapsto \bar{F}(1,t)$. Note that $\bar{\gamma}_0 = \bar{\gamma}_1$.
Now lift these maps :
$$\tilde{\gamma}_0:\{0\}\times[0,1]\to\Bbb R$$
s.t. $\tilde{\gamma}_0(0,0) =\tilde{f}(0)$ and
$$\tilde{\gamma}_1:\{1\}\times[0,1]\to \Bbb R$$
s.t. $\tilde{\gamma}_1(1,0) = \tilde{f}(1)$.
Then $k = \tilde{\gamma}_0(0,1)$. Now, $\deg f= F(1,0) - F(0,0) = \tilde{\gamma}_1(0)-\tilde{\gamma}_0(0)$, $\deg g= F(1,1)-F(0,1) =\tilde{\gamma}_1(1)-\tilde{\gamma}_0(1)$. Since $\bar{\gamma}_0 =\bar{\gamma}_1$, their lifts  $\tilde{\gamma}_0(t) =\tilde{\gamma}_1(t)-c$ for some constant $c$. Hence, $\deg f = \deg g = c$.
Converse is also true but easier.
A: For the edit, the converse is not true in general (it is however true for maps $f,g: S^n \to S^n)$.  For example, consider the $n$-torus ($n \geq 2$).  Now, take any $n \times n$ matrix $M$ with integer coefficients.  This induces a map $T^n \to T^n$ with degree given by the determinant of $M$.  In particular, matrices aren't determined by their determinant.
A: I was able to complete the proof on my own using a very neat continuity argument. Instead of providing all details, I will outline the path that took me to the answer - future readers can fill in the gaps.
Suppose $F:S^1\times I\to S^1$ is a homotopy between $f$ and $g$. Define $F': I\times I\to S^1$ as $(s,t)\mapsto (p_0(s),t)$ where $p_0 = p\vert_I:I\to S^1$. By the homotopy lifting lemma, there is a homotopy $\widetilde{F'}:I\times I\to \mathbb R$ such that the following diagram commutes:

Note that $t\mapsto \widetilde{F'}(t,0)$ and $t\mapsto \widetilde{F'}(t,1)$ are lifts of $f'$ and $g'$ respectively. By definition, $\deg f= \widetilde{F'}(1,0) - \widetilde{F'}(0,0)$ and $\deg g = \widetilde{F'}(1,1) - \widetilde{F'}(0,1)$. Define $H:I\to\mathbb Z$ by $H(s) = \widetilde{F'}(1,s) - \widetilde{F'}(0,s)$. See that $H$ is continuous. Since $I$ is connected, $H(I) \subset \mathbb Z$ must be connected, and therefore a singleton. Writing $H(0) = H(1)$ gives $\deg f = \deg g$.
