If there exists a homotopy $f \simeq g$, does this imply that a reverse homotopy $g \simeq f$ exists too, or this is not true in general?
Example: a homotopy from $\mathbb{R}^2$ to origin point $(0, 0)$.
A direct homotopy from identity map $\mathbb{1}_{\mathbb{R}^2}$ to $p_0: x \mapsto 0$, which sends every point $x$ in $\mathbb{R}^2$ to $0$, exists: $H_{\rightarrow}(t) = t\cdot p_0 + (1-t)\cdot\mathbb{1}_{\mathbb{R}^2}$.
Is it ok to claim that $H_{\leftarrow}(t) = (1-t)\cdot p_0 + t\cdot\mathbb{1}_{\mathbb{R}^2}$ is a reverse homotopy?
A homotopy from $f$ to $g$ is a continuous map $H:X\times[0,1]\to Y$ with $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Then one defines a "reverse homotopy" $\tilde H$ with $$\tilde H(x,t)=H(x,1-t)$$ from $g$ to $f$.