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Section 5.10 Inverse and direct image

A map \(f \colon X \to Y\) induces a map

\begin{equation*} f^{\ast} \colon \PP(Y) \to \PP(X) \end{equation*}

called the inverse image and a map

\begin{equation*} f_{\ast} \colon \PP(X) \to \PP(Y) \end{equation*}

called the direct image. The inverse image of a subset \(B\subseteq Y\) is the set

\begin{equation*} f^{\ast}(B) \coloneq \left\{x\in X : f(x)\in B\right\}\text{,} \end{equation*}

and the direct image of a subset \(A\subseteq X\) is the set

\begin{equation*} f_{\ast}(A) \coloneq \left\{y\in Y : (\exists x\in A)(y=f(x)) \right\}\text{.} \end{equation*}

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

\begin{equation*} A \subseteq f^{\ast}(f_{\ast}(A)) \andeq f_{\ast}(f^{\ast}(B)) \subseteq B\text{.} \end{equation*}

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:

\begin{align*} f^{\ast}\left(\bigcup_{a\in A}U(a)\right) \amp = \bigcup_{a\in A}f^{\ast}(U(a)) ;\\ f^{\ast}\left(\bigcap_{a\in A}U(a)\right) \amp = \bigcap_{a\in A}f^{\ast}(U(a)) ;\\ f^{\ast}(\complement U) \amp = \complement(f^{\ast}(U)) \text{.} \end{align*}

For the direct image, one only has

\begin{align*} f_{\ast}\left(\bigcup_{a\in A}U(a)\right) \amp = \bigcup_{a\in A}f_{\ast}(U(a)) ;\\ f_{\ast}\left(\bigcap_{a\in A}U(a)\right) \amp \subseteq \bigcap_{a\in A}f_{\ast}(U(a)) \text{.} \end{align*}

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

\begin{equation*} X = \bigcup_{y \in Y} f^{\ast}\{y\}\text{.} \end{equation*}

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!!