How can $\text{deg }(f)= \text{d }(f) $ if they're defined differently?

Question

From Rotman's Algebraic Topology:

Prove: If $f : S^1 \rightarrow S^1, $ then $\text{deg }f = d(f)$.

But based off of the definitions Rotman provided below, I do not see how this is possible:

"$\text{deg }$" is defined as: If $f : (I, \dot I) \rightarrow (S^1,1)$ is continuous with $I = [0,1]$ and $\dot I = \{0,1\}$, define the degree of $f$ by $\text{deg } f = \tilde f(1)$, where $\tilde f$ is the unique lifing of $f$ with $\tilde f(0) = 0$.

"$d$" is defined as: A continuous map $f : S^n \rightarrow S^n$ (where $n \gt 0$) has degree $m$, denoted $d(f) = m$, if $f_* : H_n(S^n) \rightarrow H_n(S^n)$ is multiplication by $m$.

The proof uses the Hurewics isomorphism $\psi$ and a theorem:

If $f$ is a closed path in $S^1$ at $1$ and if $m \in \Bbb Z$, then $t \mapsto f(t)^m$ is a closed path in $S^1$ at $1$ and $\text{deg }f = m*\text{deg }(f)$.

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But how can this be possible? If we're assuming $f : S^1 \rightarrow S^1$, then $\text{deg }f$ is undefined.

Answer 1

As @JustinYoung observed this is Exercise 4.16 and as @CaptainLama said maps involved are canonically the same.

I will elaborate on this idea in the context of mentioned exercise.

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We start from a map of pointed spaces $f\colon (S^1, 1)\to (S^1, 1)$.

Degree via homology The map induces a map on homology $f_*\colon H_1(S^1)\to H_1(S^1)$. But $H_1(S^1)=\mathbb Z\langle g\rangle$, so $f_*$ is multiplication by some integer called $\deg_H f\in \mathbb Z$: $$(\deg_H f) \cdot g = f_*(g).$$

Degree via fundamental groups We have a closed path $h\colon (I,\dot I) \to (S^1, 1)$ given by $h(t)=\exp 2\pi i t$. We define a closed path $f'\colon (I, \dot I)\to (S^1, 1)$ by composition

$$f' := f\circ h$$

We have a covering space $p\colon \mathbb R\to S^1$ where $p(t)=\exp 2\pi i t$. Now we can lift path $f'$ to the unique path $\tilde f\colon I\to \mathbb R$ such that $\tilde f(0)=0$. Then we define the degree of $f$ to be $$ \deg_{FG} f = \tilde f(1) $$ which is necessarily an integer, what comes from the fact that the path $\tilde f$ is closed and $p^{-1}(1)=\mathbb Z$.

The claim

$$\deg_H f = \deg_{FG} f$$

Proof: Starting from a map $f$ with $n=\deg_{FG} f$ we homotope it to one of the standard paths $$a_n\colon (S^1, 1)\to (S^1, 1), \quad a_n(z) = z^n.$$ Observe that $f_*=(a_n)_*$, hence $$\deg_H f = \deg_H a_n.$$ Moreover using the homotopy from $f$ to $a_n$ you can construct a homotopy from $f'$ to $a_n'$. This implies $$\deg_{FG} f = \deg_{FG} a_n.$$

Hence it suffices to prove the claim only for "standard" functions $a_n$ which you can do using the Hurewicz theorem.

Answer 2

I'll clarify what I meant: there is a theorem in the book that the map $\pi_1(S^1, 1)\to \mathbb Z$ given by deg is an isomorphism. So there is a unique homotopy class of paths $i$ so that $deg(i) = 1$, then my claim is that you can define the degree of a (let's say pointed) map $f: S^1 \to S^1$ by looking at $deg(f\circ i)$, this is my best attempt to get what the book means by defining the degree of $f$ "in terms of the fundamental group". If you define degree this way, the naturality of the Hurewicz map yields the result.