# Fibration factorization leads to new fibrations.

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.

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What is $W$? How exactly you would start then, and what exactly we are about to prove? Can you briefly share us the injective case? Have you used the condition that $t\circ s$ is a fibration? –  Berci Nov 8 '12 at 23:23
To prove b) I want to use the condition that $f$ is a fibration. But to use it, I need $W\times I \to X \to Z$ to be a map, which is my problem. $W$ is the arbitrary space out of the Hurewicz fibration diagram. –  x x Nov 9 '12 at 10:36