I've got a map between two infinite dimensional spaces, $f: A\to B$, where $A$ seems "larger" than $B$. For the sake of conversation let's assume that $A$ is the set of smooth maps $\mathbb R^3\to \mathbb R^3$ and $B$ is the set of smooth maps $\mathbb R^2\to \mathbb R^2$. I would like to figure out whether $B$ has a point $p$ such that $|f^{-1}(p)|>1$.


If $A,B$ were finite sets with $|A|>|B|$ the conclusion would follow from the pigeonhole principle.


If $A,B$ were finite dimensional spaces with $\dim(A)>\dim(B)$ the conclusion would follow as well. 


The above notions of "size" of finite sets and finite dimensional spaces, the number of elements and the number of elements in the linear basis, are very useful because in part because they are finite and also because one can make "size" based arguments such as pigeonhole principle regarding the corresponding maps.


What could be an appropriate "size" for infinite dimensional spaces such as smooth maps?



What should be considered a finite size of an infinite dimensional space?
I don't think you'll find a good notion of dimension to get the pigeonhole principle to work.


Here are two examples of injective maps that are visualizable. I've given formulas for concreteness, which may or may not help in the visualization. They use $C(X,Y)$, the set of continuous maps from $X$ to $Y$. 


1) $\ A = C(S^1,R^3), \ B=C(S^1,R^2)$. We want to show that given a curve $a$ in $R^3$, there is a curve in $R^2$ that has all its information. We do this by finding a curve whose first half gives the $x$-coordinate of $a$, and whose second half gives the $y$ and $z$ coordinates.


If $a=(a\_x, a\_y, a\_z)$, let $h = a\_y(0)-a\_x(0)$, and let $f(a)$ be


$$f(a)(t) = (a\_x(2t) + h,\ 
a\_z(0) + (2t-1)th) \text{ if } t \le 1/2$$ 


$$f(a)(t) = (a\_y(2t-1),\ a\_z(2t-1) \ \ \ \ \ \ \ \ \ \ \ ) \text{ if } t \ge 1/2$$


Then $f(a)$ is continuous, and $f$ is injective.


2) $\ A' = C(S^2,R^3), \ B=C(S^1,R^2)$. We want to show that given an image of a sphere in $R^3$, there is a curve in $R^2$ that has all its information. We do this by finding a curve that gives us the information meridian by meridian.


Suppose $s$ is in $A'$. For any integer $i$, let $s\_i$ be the image under $s$ of the meridian at $i$ radians. Then $s\_i$ is in $A$. For any $x\in[0,1]$ and $i\in N$, let
$$g(s)(2^{-(i+1)}(1+x)) = (1-2^{-i})\,f(s\_i)(0) + 2^{-i}f(s\_i)(x)$$
$$g(s)(0) = f(s\_0)(0)$$


On each interval $[2^{-(i+1)}, 2^{-i}]$, $g$ reproduces $f(s\_i)$, but smaller and faster so as to be continuous. Since $g(s)$ lets us reconstruct $s$ on a dense set of meridians, and since $A'$ had only continuous maps, we can reconstruct $s$ fully. So $g$ is injective.


Thus we have injective maps from $A$ to $B$ and from $A'$ to $B$. If no notion of dimension will yield a pigeonhole principle for examples as nice as these, then probably no notion of dimension will work at all.


