# Existence of a structure-preserving mapping between two spaces?

I have some questions, but not sure if they are meaningful:

1. Suppose $X$ and $Y$ are two arbitrary measurable spaces. Does there exist a measurable mapping from $X$ to $Y$?
2. Suppose $X$ and $Y$ are two arbitrary measure spaces. Does there exist a measure-preserving mapping from $X$ to $Y$?
3. Suppose $X$ and $Y$ are two arbitrary topological spaces. Does there exist a continuous mapping from $X$ to $Y$?

They are similar in this way:

if $X$ and $Y$ are two arbitrary sets with some type of structure, does there exist a structure-preserving mapping from $X$ to $Y$?

Hope to see if there can be some insights.

Thanks and regards!

-
Every constant map $f:X\to Y$ is measurable and continuous (wrt. to every $\sigma$-algebra over $Y$ and every topology over $Y$, respectively), so answers to the first and third questions are positive. The answer to the second question is negative -- consider the measure space $Y$ with the zero measure and $X$ with a nonzero measure. – Damian Sobota Nov 25 '11 at 1:25
Thanks! How about non constant mapping? – Tim Nov 25 '11 at 1:26
Re measurability, all I can think is that if the map is differentiable, then the measure is preserved precisely when the determinant of the Jacobian is 1. since a mapping f between abstract measurable spaces $(X,\mu,\sigma), (Y,\phi, \sigma'$ is measurable if/when $f^{-1}(U)$=V, where U is in \sigma and V is in sigma', you can always construct a measurable map f between them, by demanding than an element in sigma' be sent to an element in sigma under $f^{-1}$. Same goes for topological – gary Nov 25 '11 at 1:37
(cont.)spaces A,B, whereyou can just construct g:A-->B , so that $g^{-1}(W)=Z$ , for any W open in Y, and some Z open in X. I think this is part of what pointless topology is about. Is that your question? – gary Nov 25 '11 at 1:37
@Tim: if $X$ is connected with cardinality greater than 1 and $Y$ is discrete, then there is no continuous map from $X$ to $Y$ other than a constant one. – Damian Sobota Nov 25 '11 at 1:41

I gather in an answer what has been said in the comments to the starting post.

The most important issue concerning the boxed question is that it is too general. As we showed in comments an answer strongly depends on the structure we consider and such a general statement allows us to construct completely trivial answers. This answers can be either positive or negative depending whether we put additional conditions the structure has to fulfill or not.

For example, consider the category of topological spaces with morphisms being continuous maps. The question is: given two topological spaces $X$ and $Y$ is there a map $f:X\to Y$? In this general statement the answer is positive with the argument: every constant function is continuous. Ok, what about nonconstant maps? If the space $X$ is connected, $|X|>1$, and $Y$ is totally disconnected (only singletons are connected, eg. $Y$ is the Cantor set), then such a map does not exist, because continuous maps preserve connectedness. So again we can add some additional conditions to the questions, for example we can consider the maps which are embeddings (ie. homeomorphisms onto the image). And again the answer is negative, for we cannot embed the circle $S^1$ into the segment $[0,1]$.

The same concerns algebraic structures. For example, take the category of vector spaces with linear maps as morphisms. For vector spaces $V$ and $W$, there is always a zero map $z:V\to W$ which is of course linear. But this gives us nothing. So let us constrain the linear maps only to injections. Then the answer is negative, because if $\dim V>\dim W$, then there is no injection of $V$ into $W$.

Similarly, if we consider category of fields and maps being homomorphisms, then for two fields of different characteristics there is no homomorphism between them in case $1$ is added to the structure and there is always a zero homomorphism if we remove $1$ from the structure.

In case of measure spaces the answer is similar to the above ones. Let $(X,\mu)$ and $(Y,\nu)$ be two measure spaces. If $\mu(X)>\nu(Y)$, then obviously there is no measurable measure preserving function between $X$ and $Y$. But if $X\subseteq Y$ and $\mu=\nu$, then the identity (the inclusion) simply preserves $\mu$.

And there are many, many other examples like mentioned by Arturo Magidin the category of posets or even the simple category of sets and any functions as morphisms showing that the answer strongly depends on the structure we consider.

-