Measurable maps

1. Sep 23, 2007

P3X-018

I have 2 maps f and h such

$$f :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Y}, \mathbb{K})$$
$$h :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Z}, \mathbb{G})$$

where $\mathbb{K}$ and $\mathbb{G}$ are $\sigma$-algebras on the spaces Y and Z respectively, and $\mathbb{E} = \sigma(f)$ is the $\sigma$-algebra generated by the map f.
f is assumed to be surjective and $\mathbb{G}$ is assumed to separate points in Z. That is for any 2 different point a and b in Z we can find a set $A\in \mathbb{G}$ such that $a\in A$ and $b \notin A$.

The problem is that, assuming h is $\mathbb{E}$-$\mathbb{G}$ measurable, to show that there is a map $g:\, \mathcal{Y} \rightarrow \mathcal{Z}$ such that $h = g\circ f$.

I don't even know how to attack this problem. What is even meant by showing that there IS such a map? A hint could be helpful.

Last edited: Sep 23, 2007
2. Sep 23, 2007

EnumaElish

Can you go from (Y,K) to (X,E)? If you can, then you know how to go from (X,E) to (Z,G).

3. Sep 23, 2007

P3X-018

So if I could go from (Y,K) to (X,E) that is if there was a inverse map of f, that would be enough to conclude that THERE IS a map g?
But f would have to be bijective inorder to go from Y -> X.
Can f be injective under the given assumption? Because it's already surjective, so if it were injective we would be done.

4. Sep 23, 2007

matt grime

You do not assume injectivity.

Clearly the hypotheses have been given for a reason. So start to see what you can deduce from them. Let's see what you need to do. You need, given a point y in Y to find a way to associate it to a point g(z) in Z. And we need to do it in such a way that gf(x)=h(x). Wow, what do you know, from just writing that down I've essentially solved the question for you....

5. Sep 23, 2007

P3X-018

I didn't mean that I assumed f to be injective, only that wether it should (or could) be shown that f is injective, and if so would that solve the problem, since we could go from Y to Z by h(f^(-1)(y)).
Hmm this sounds too 'simple'...
So basically the argument is, that since f is surjective then for every y in Y there is a x in X, such that y = f(x), and since z = h(x), z is the element in Z that is mapped into by y = f(x), with g(f(x)) = h(x), that is z is mapped by z = g(y).

EDIT:
Another question is to show that g is K-G measurable.
Now if I could look at the map from Y -> X that is $f^{-1}$, then I could use the theorem that says that g is K-G measurable if and only if $f^{-1}$ is K-E measurable, assuming that h is E-G measurable.
I'm pretty certain of that I need to use this theorem, somehow, to show the measurability of g, since obviously I can't investigate wether $g^{-1}(A) \in \mathbb{K},\,\, \forall A\in\mathbb{G}$.

Last edited: Sep 23, 2007
6. Sep 24, 2007

matt grime

What was the other assumption in the original question that you've not used? The one abotu G separating points. PErhaps that is useful.