Partition into "fibers" $f^{-1}(y) \in Y$

Question

Consider any surjective map f from a set X onto another set Y. We can define an equivalence relation on X by $x_1Rx_2$ if $ f(x_1)=f(x_2)$. Check that this is an equivalence relation. Show that the associated partition of X is the partition into "fibers" $f^{-1}(y) \in Y$.

I know how to do the first part. My question is about the second part: "Show that the associated partition of X is the partition into "fibers" $f^{-1}(y) \in Y$"

I know what a partition is, but what does this mean? What is the "partition into fibers"?

Thanks!

Answer 1

As James points out, the fiber over an element $y \in Y$ is the subset $f^{-1}(y) = \{x \in X \colon f(x) = y\} \subset X$.

For different elements $y \in Y$, the subsets $f^{-1}(y) \subset X$ are disjoint (do you see why?). Moreover, every element of $X$ belongs to some fiber $f^{-1}(y)$ (again: do you see why?). Thus, the set $X$ is partitioned into subsets (these subsets being the fibers).

On the other hand, the set $X$ has an equivalence relation $R$, which you describe above. The equivalence classes of $R$ also determine a partition of the set $X$.

You have to show that these two partitions of $X$ are actually the same.

Answer 2

The fibers are just the preimages in $X$. So for example, if $f(x) = y$ then $x$ is in the fiber of $y$, i.e. $f^{-1}(y)$ is the set of all $x \in X$ that get sent to $y$. So I think your notation should be $f^{-1}(y) \in X$ and not $\in Y$.

As far as the term "fiber" goes, I just think of the projection $\pi_1: \mathbb{R}^2 \to \mathbb{R}$ where $(x,y) \mapsto x$. So the fiber of each point on the real line actually looks like a fiber going up and down through the plane.