Section 5.10 Inverse and direct image
A map \(f \colon X \to Y\) induces a map
called the inverse image and a map
called the direct image. The inverse image of a subset \(B\subseteq Y\) is the set
and the direct image of a subset \(A\subseteq X\) is the set
These operations are related in the following manner: one has \(f_{\ast}(A)\subseteq B\) if and only if \(A\subseteq f^{\ast}(B)\text{.}\) In particular, we have
In general, both of these containments are strict. However, if \(f\) is an injection, then \(A = f^{\ast}(f_{\ast}(A))\text{,}\) and if \(f\) is a surjection, then \(f_{\ast}(f^{\ast}(B)) = B\text{.}\)
In many respects, the inverse image is more natural than the direct image. For example, the inverse image preserves unions, intersections, and complements, so that one has:
For the direct image, one only has
There is no containment between \(\complement f_{\ast}(A)\) and \(f_{\ast}(\complement A)\) in general.
In the particular case where \(B = \{y\}\text{,}\) the inverse image \(f^{\ast}\{y\}\) is called the fiber of \(f\) over \(y\text{.}\) This gives us a handy way to think about maps \(f \colon X \to Y\text{:}\) in effect, they organize \(X\) into a disjoint union of fibers. That is, if \(y \neq y'\text{,}\) then \(f^{\ast}\{y\} \cap f^{\ast}\{y'\} = \varnothing\text{,}\) and
The map \(f\) is injective if and only if each fiber \(f^{\ast} \{y\}\) has at most one element, and \(f\) is surjective if and only if each fiber \(f^{\ast} \{y\}\) has at least one element.
Warning 5.10.1.
Now for the annoying news. The notations \(f^{\ast}\) and \(f_{\ast}\) are not the usual notations. The more typical notation for the inverse image of \(B\) is \(f^{-1}(B)\text{.}\) The more typical notation for the direct image of \(A\) is \(f(A)\text{.}\) This notation makes it look as though these operations are inverse — in general they are not!!