let $f\colon X \to Y$ be a fibration. Let be $s\colon X \to Z$ and $t\colon Z \to Y$ be maps, such that $t\circ s = f$. I have to show: a) If $t$ is injective then $s$ is a fibration, b) if $s$ surjective then $t$ is a fibration.
a) is clear, with b) I am not sure:
We have the diagram $W\times{0} \to Z \to Y$ commuting with $W\times{0} \to W\times I \to Y$, we have to find a map $W\times I \to Z$ so that all is commuting.
Now I should expand the first chain of maps with $X$ by taking preimages of the surjective $s$ map to $W\times{0} \to Z \to X \to Y$, so that I can use the fibration property of $f$ giving a map $W\times I \to X$ so that all is commuting. This map could be sent to $W\times I \to X \to Z$ by composing $s$, which I wanted to find.
But I don't see why the function $W\times{0} \to Z \to X$ should be a map, the function inverse $s$ $Z \to X$ being a map would be sufficient.
If $s$ would be an open map, this would be the case, but it only is said to be a surjective map.
What am I overseeing?
a) goes like this:
We have the diagram $W\times{0} \to X \to W$ commuting with $W\times{0} \to W\times I \to W$, we have to find a map $W\times I \to X$ so that all is commuting.
Composing both map chains with $t$ gives a diagram $W\times{0} \to X \to W \to Y$ commuting with $W\times{0} \to W\times I \to W \to Y$ where $X \to W \to Y$ is $f$.
Now I use the fibration property of $f$ and gain the map $W\times I \to X$ commuting nicely with the previous diagram.
Because $W\times I \to X \to W \to Y$ and $W\times I \to W \to Y$ are commuting and $t$ is injective, I can remove $t$ and $W\times I \to X \to W$ and $W\times I \to W$ still are commuting, so that $s$ is a fibration.