Fundamental group of composition of function

Question

Let Y be a simply connected space, and let f : X → Y and g : Y → Z be continuous functions. What is π1(g ◦ f )? Prove your answer.

I think the fundamental group of the composition would be the trivial group. I am not sure how to prove this!!

Any hints?

Answer 1

Since $\pi_1$ is a functor, we have $\pi_1(g\circ f)=\pi_1(g)\circ \pi_1(f)$. However, $Y$ is simply connected, meaning that $\pi_1(Y)=c_{y_0}$, the constant path at the point $y_0$. Thus $g$ is homotopic to a map that sends $Y$ to a point, say $z_0$, in $Z$. Therefore, the homomorphism $\pi_1(g\circ f):\pi_1(X)\rightarrow \pi_1(Z)$ is the trivial homomorphism sending all the elements of the group $\pi_1(X)$ to the neutral element of $\pi_1(Z)$, say $c_{z_0}$.

EDIT : I used that $\pi_1$ is homotopy invariant, which is your previous question ;)