let $S^1$ be the unit circle of complex plane and let $f, g:S^1 \to S^1$ the map $f(z)=z$ and $g(z)=z^2$
Is $f$ homotopic to $g?$
My attempt : Yes, I think
Take the map $F : S^1 \times [0,1] \to S^1$ defined by $F(e^{i\theta},t)=e^{i(\theta + t\pi) }$
Also, $F$ is continious because it is the compoisition of map
$S^1 \times I \to S^1 \times S^1 \to S^1 \to S^1$
Your $F$ is a homotopy between $f$ and $h(z):=-iz$. Since $h$ is just a rotation of $f$ by $\pi$, they are clearly homotopic.
You may need some firepower to do this. For instance you can define the notion of degree, and show that it's a homotopy invariant. Since $f$ has degree $1$, $g$ degree $2$, they are not homotopic.
I believe that the name Heinz Hopf comes up here, around the beginning of the $20$th century.