Fundamental Group equaling 0

Question

Let $X$ be a space for which $\pi(X,x)=0$. If $f,g$ are two paths in $X$ with $f(0)=g(0)=x$ and $f(1)=g(1)$, why is $f$ equivalent to $g$?

Answer 1

Since $f(1)=g(1)$, it follows $f*\bar{g}$ is a loop at $x$. Thus $[f*\bar{g}]\in\pi_1(X,x)$, which is trivial by assumption, so $[f*\bar{g}]=[e_x]$, that is, $f*\bar{g}$ is path-homotopic to the constant loop $e_x$ at $x$. Then you have $$ [f*\bar{g}]*[g]=[e_x]*[g]\implies [f]=[g], $$ so $f$ and $g$ are path-homotopic.