@0celo7 it's more just a special case of the $H : [0,1] \times X \rightarrow X | (t,x) \mapsto H(t,x)$ homotopy map in
en.wikipedia.org/wiki/Homotopy but instead of $H(0,x) = f(x)$, $H(1,x) = g(x)$. The intuition is to picture $f$ as the top arc of the unit circle, $H(0,x) = f(x) = \sqrt{1-x^2}$, $g$ as the bottom arc $H(1,x) = - \sqrt{1-x^2}$ and $H(t,x)$ as any curve inside the circle. You picture $H$ as just continuously deforming the top arc into the bottom arc, it's very natural.