If $f$ is nullhomotopic, then any $g\circ f$ is too

Question

If $f:X\to Y$ is nullhomotopic, then any composition, with $g:Y\to Z$, has that $g\circ f$ is nullhomotopic.

Proof:

1. if $f:X\to Y$ is nullhomotopic, then there is a homotopy $f_s:X\to Y$ where $f_0=f$ and $f_1=c_y$, where $c_y(x)=y$ for all $x\in X$ and $y\in Y$.
2. If $f\cong c_y$, then $g\circ f\cong g\circ c_y$.
3. $g\circ c_y(x)=g(y)$ for every $x\in X$, so that $g\circ f\cong c_{g(y)}$

Is this ok?

Answer 1

Looks fine. Another approach: a map is nullhomotopic if and only if it factors through a contractible space. If f factors, then so does gf.