If we have two topological spaces X,Y and a continuous function between f, and $f(a)=b$.
We get a function between the fundamental groups:
$f_*: \pi(X,a)\rightarrow\pi(Y,b)$, which is given by $f_*([g])=[f \circ g]$. But it is also so that f must be both surjective and injective for $f_*$ to be injective?
If you get have that $[f\circ g_1]=[f\circ g_2]$, can you please tell me why we do not get that $[g_1]=[g_2]$ if we only have that f is injective, we need surjectivity aswell?
PS: My question comes by wondering about the classical example from $S^1\rightarrow B^2$, if we just use the identity we get that f is injective, but not surjective, and it is proved that the fundamental group of the last one is trivial and the first one is not. So $f_*$ is not injective. But I do not understand where the details go wrong.