# If $q\circ p$ is a normal covering map then $p$ also is a normal covering map.

Take $p:F\to E,q:E\to X$ maps such that $q$ and $q\circ p$ are covering maps. If $X$ is locally path-connected then we know that $p$ is a covering map.

I want to prove that if $q\circ p$ is normal then $p$ is normal.

For this, I took $e\in E$, $x=q(e)$ and $f,f'\in p^{-1}(e)$. Since $f,f'\in (q\circ p)^{-1}(x)$, then there is a $X-$Deck transformation $\sigma$ of $F$ such that $\sigma(f)=f'$, i.e. a homeomorphism of $F$ such that $(q\circ p)\circ\sigma=q\circ p$.

I want to prove that in this case $\sigma$ can also be seen as a $E-$Deck transformation.

By the way, if we suppose all the spaces are path-connected the problem becomes very easier: If $K,H,G$ are the fundamental groups of $F,E,X$ respectively, then the problem reduces to prove that $K\trianglelefteq H$ given that $K\trianglelefteq G$ (with respect to the inclusions of the fundamental groups induced by the covering maps), which is almost trivial.