Suppose $h:S^1 \to S^1$ is nullhomotopic, and $f:S^1 \to S^1$ defined by $f(x)=-x$ Then $f\circ h$ is nullhomotopic.
I've seen this in a proof, And I want to convince myself by coming up with an explicit homotopy but I am struggling. Any help?
Take Say $H:S^1\times I \to S^1$ to be the Homotopy between $h$ and the constant function $c$. with $$H(s,0)=h(s) $$ $$H(s,1)=c$$ I need to define a homotopy $F:S^1\times I \to S^1$ most likely in terns of $H(x,t)$ between $f \circ h$ and $c$.
Recall that if $f$ and $g$ are homotopic via $H$, then $k \circ f$ and $k\circ g$ are homotopic via $k\circ H$.
Given that $h: S^1 \to S^1$ is nulhomotopic, let $H$ be the homotopy between $h$ and the constant map $g(x) = c \ \ \forall x \in S^1$ for some $c \in S^1$.
Now, $f: S^1 \to S^1$ is defined as $f(x) = -x$, then, $f \circ H$ is a homotopy between $f\circ h$ and $f \circ g$. But, since $g$ is constant map, so is $f\circ g$. Thus, $f\circ H = -H$ serves as the required homotopy between $f\circ h$ and constant map $f \circ g$.