$f$ is null homotopy if and only if $f_*$ is the trivial induced homomorphism

Question

I want to show that $f:S^1\to S^1$ is homotopic to the constant map if and only if $f_*:\pi_1(S^1,x_0)\to \pi_1(S^1,x)$ , $f_*([\gamma])=0$ This seems like it should be an obvious fact but I am having trouble with details.

One direction is obvious. That is, since $f\sim g\implies f_*=g_*$, if I assume $f\sim c$ for $c$ constant map, then $f_*=c_*=0$.

The other direction is suppose $f_*=c_*$ for constant function $c$. I want to show that $f\sim c$.

Answer 1

If $f$ omits any point of $S^1$, the image of $f$ is contractible. In particular, if $f$ is constant, $f(S^1)$ is contractible and $f_*$ is the trivial map.

The other direction is false. Let $S^1$ be represented by the set $\{(x,y) \in \Bbb{R}^2 \mid x^2 + y^2 = 1\}$. Let $f(x,y) = (|x|,y)$ and $x_0 = (1,0)$. Then $f$ is continuous, nonconstant, and maps every element of $\pi_1(S^1,x_0)$ into a contractible set, the right semicircle, so $f_*([\gamma]) = 0$ for all $[\gamma] \in \pi_1(S^1,x_0)$.

(If one thinks a covering space argument might work... Consider $f(\mathrm{e}^{\mathrm{i}\hat{\theta}}) = \mathrm{e}^{\mathrm{i}\arctan(\hat{\theta})}$, where now $\hat{\theta}$ ranges over $\Bbb{R}$ and every $2\pi$ the point is on a different sheet of the covering. This maps the entire covering space to the same semicircle.)

Answer 2

It's important to distinguish between being equal to the constant map and being homotopic to the constant map. Eric Towers' answer gives you a map that induces the zero map on $\pi_1$ but is not the constant map. However, it will be homotopic to a constant map.

Doing the other way round: you seem to already be aware that $\pi_1(S^1, x_0) \cong \mathbb Z$, generated by a loop that you call $\gamma$. So let $f: S^1 \to S^1$ be any continuous map. Choose a basepoint $x_0 \in S^1$.

Hopefully you agree that we can take $f$ to be basepoint preserving: rotating the second copy of $S^1$ so that $f(x_0) = x_0$ won't change whether or not $f$ is homotopic to a constant map, as we can start any homotopy with a rotation undoing the first one.

What do we mean by the map $f_*: \pi_1(S^1, x_0) \to \pi_1(S^1, x_0)$? It's determined by where it sends $[\gamma]$, and $f_*([\gamma]) = [f \circ \gamma]$. But since $f: S^1 \to S^1$ and $\gamma$ is the identity on $S^1$, this means that $f_*([\gamma]) = [f]$. In particular, if $f = n\gamma$ then $f_*([\gamma]) = [n\gamma]$

Note that since we chose $f$ to be basepoint preserving, it gives us an element of $\pi_1(S^1, x_0)$, and so we know that $[f] = n[\gamma]$ for some $n$. Therefore $f_*([\gamma]) = (n\gamma)_*([\gamma]) = [n\gamma]$, which is zero iff $n$ is zero iff $f$ is homotopic to a constant map.