# A question on degrees of maps of the fundamental group of the unit circle

1. Jan 6, 2013

### trmukerji14

Hello,

I'm reading a textbook and in the textbook we are discussing the fundamental group of the unit circle and having some difficulty making out what a degree of a map is and why when there is a homotopy between two continuous maps f,g from S$^{1}$ to S$^{1}$ why the deg(f)=deg(g)

We have that the fundamental group of the unit circle, $\pi _{1}$(S$^{1}$ )

f,g: S$^{1}$ →S$^{1}$ are continuous and f≈g(homotopy)
$\iota$ $\in$$\pi _{1}$(S$^{1}$ ) is a generator
The book defines the degree of f, deg(f), as the integer with respect to the composite
$\pi _{1}$(S$^{1},1$ ) $\stackrel{f_{*}}{\rightarrow}$ $\pi _{1}$(S$^{1},f(1)$ )$\stackrel{\gamma_{a}}{\rightarrow}$ $\pi _{1}$(S$^{1},1$ )

Note that a is a path from 1 to f(1) and $\gamma_{a}$ is the change of base-point isomorphism

We have $\iota$→deg(f)$\iota$

What exactly is this degree of f?

My understanding is that when we consider $\iota$ the associated integer is 1 and then we have that the associated integer with $\pi _{1}$(S$^{1}, f(1)$ )

What does the change of basepoint homomorphism have to do with it?

Also, why does deg(f)=deg(g)? I know this is due to the abelian property of $\pi _{1}$(S$^{1},1$ ) and the fact that there is a path between f(1) and g(1).

Any help on this would be appreciated greatly.

Thank you