Badly explained solution My algebraic topology class is very bad at teaching, it just doesn't explain what's needed. Let me be specific, I am looking at this question,

Q. Find the degree of $f_0 :S^1 \to S^1$ the constant map, such that $f_0: z \to 1$

Now, it spends a great deal of time and space on paper explaining how "degree" is defined here.(please skip if you already know all of this)

We take the purely topological definition of the degree of a map $f:S^1 \to S^1$  in terms of the number of path components of a associated non-compact space $f^*\mathbb{R}$(pullback of $f$).
We define $p: \mathbb{R} \to S^1; t \to e^{2 \pi i t}$ and the pullback of $p$ along the map $f:X \to S^1$ is the subspace of $X \times \mathbb{R}$ defined by
$$f^*\mathbb{R}=\{(x,t) \in X \times \mathbb{R}: f(x)=p(t) \in S^1\}$$
The absolute degree $\lvert\deg\rvert f$ of $f:S^1 \to S^1$ is defined as "the number of path components of $f^*\mathbb{R}$ if it is finite" and $0$ when infinite
The sign of the degree of $f$ is defined using
$$z^{-1}f: S^1 \to S^1; z \to z^{-1}f(z)$$
to be the sign of the jump in the absolute degrees in going from $f$ to $z^{-1}f$(you know what, I lost it here. what "signs"? Plus or minus? How does that relate to this map? and what "jumps"? what is a jump here? Explodes my brain with abstractness)
$$\epsilon(f)=\lvert\deg\rvert(f)-\lvert\deg\rvert(z^{-1}f) \in \{-1,1\}$$
The degree of $f:S^1 \to S^1$ is defined by
$$\deg(f)= \epsilon(f)\lvert\deg\rvert(f) \in \mathbb{Z}$$

I thought I'd just do the question because I figured it'd be faster and easier to learn than just looking at these indescriptive words.
well, the solution has what seems like a completely irrelevant approach. It talks about homeomorphisms and doesn't talk about this $\epsilon$ thing.

The pull back $f^*_0\mathbb{R}$ is such that there is a defined homeomorphism
$$S^1 \times \mathbb{Z} \to f_0^*\mathbb{R};(z,n) \to (z,n)$$
as $S^1$ is path connected and $\mathbb{Z}$ has an infinite number of path components $f_0$ has degree $0$. The absolute degree is $0$ in which case the degree is $0$.

Well, what was the point of going through all the tedious definitions? It doesn't use the $\epsilon$ to find the degree and talks about this homeomorphism instead. I managed to convince myself that $S^1$ is path connected, but what if $\mathbb{Z}$ has infinite path components? I thought the pullback is of $\mathbb{R}$ and not $\mathbb{Z}$ why is it talking about the path components of $\mathbb{Z}$?
I really don't understand this, and very stressed about how this all works. Does anyone know how to clear this up?
 A: The degree of a map $S^1\to S^1$ has the following interpretation : up to deformation (homotopy), these maps are all equivalent for some $n\in \mathbb{Z}$ to the map that twists the circle $n$ full times around itself (the sign of $n$ tells you if you twist left or right, or more correctly in the direct or the indirect sense). Such a map can be described as $z\mapsto z^n$ if you view $S^1$ as the unitary complex numbers.
What the definition with the number of connected components tells you is the number of twists of $f$, but it doesn't tell you if $f$ twists directly or indirectly. So the trick here is the following : we know how to twist indirectly once, it's just $g: z\mapsto z^{-1}$. So if we do $g\circ f$, there are two possibilities : if $f$ twists in the direct sense, then composing with $g$ will decrease the number of twists ; but if $f$ twists indirectly, then composing with $g$ will increase the number of twists (it decreases the degree, but in the negatives, so it increases the absulute value of the degree). So to know the sign of the degree, you do the difference of the absolute degree of $f$ and $g\circ f$ : this is your $\epsilon$, which is $\pm1$ according to the case. And to get the degree, you take the product of the absolute degree (defined as the number of connected components etc.) and $\epsilon$.
In your example, $\epsilon$ doesn't come up because the absolute degree is $0$ (the space for which you have to count the number of connected components has an infinite number of such components). So you don't need to find out if the twist is direct or indirect : there is no twist.
