For a path connected space $X$, it is simply connected iff any two paths sharing endpoints are homotopically equivalent. Here simple connectivity means it has trivial fundamental group.
I'm doing the if part, so let's suppose $X$ is path connected with equivalence property. Now given $a\in X$, I wish to show $\pi_1(X,a)=1$. Let $f\in\Omega(X,a)$, now if $f$ consists solely of a single point we are happy, if it contains another point say $b\ne a$, then we can split $f$ into two consecutive paths $g,h$ such that $h$ follows $g$ and $f=g*h$ (note: path multiplications are to be read from left to right, unlike ordinary map composites). Now, I have $h=g^-$, so, up to path homotopy, we have $$[f]=[g][h]=[g*g^-]. $$ Now I can't figure out how to show $g*g^-$ is equivalent to the constant path at $a$, although it is intuitively evident. Also for the converse question: if $[g*h]=[c]$ ($c$ means constant path) then how to prove $h\sim g^-$, I completely have no clue either. Intuition doesn't get me anywhere.
Silly as the two questions may look, could anyone enlighten me on them? Best regards to you!