I'm reading Rotman's Algebraic Topology and the definition of degree is as given:
If $f : (I, \{0,1\}) \rightarrow (S^1, 1)$ is continuous, define the degree of $f$ by: $$\deg f = f'(1)$$ where $f'$ is the unique lifting of $f$ with $f'(0) = 0.$
What I don't understand is why $f'(1) \neq f'(0)$.
If a lifting of $f$ is defined as: $f' : (I, \{0,1\}) \rightarrow (R,t_0)$ (where $t_0 \in \Bbb Z$) is a pointed map and $\exp : (\Bbb R, t_0) \rightarrow (S^1, 1)$ and pointed maps preserve base points $\Rightarrow$ $f'(0)$ should equal $f'(1)$?
What am I missing here?
You should think about it as $(I,\{0,1\},0)$, $(S^1,\{1\},1)$ and $(\Bbb R,\Bbb Z,t_0)$.
The notation $(I,\{0,1\})$ probably just is used to suggest that $0$ and $1$ are mapped to the same point and shout not be seen as the base point. In fact by the universal property of quotient topology you know that these map are exactly the maps which have domain $S^1$. So you basically looking at self maps of $S^1$.
I would guess $(I,\{0,1\})$ is chosen in favor of $S^1$ to be in line with the covering space theory. It usually goes this way: if you have a covering and a simply connected space then a lift exists.
If you would require $\{0,1\}$ be mapped to $t_0$ then a lift does not have to exist. You could just map $I$ and $\Bbb R$ via the exponential $e^{2\pi t}$ to $S^1$. Then the lift has to be the inclusion up to $\Bbb Z$(a base point would give uniqueness). But the inclusion is injective so $f'(1)\ne f'(0)$.