TITLE: Overview of basic results about images and preimages
QUESTION [171 upvotes]: Are there some good overviews of basic facts about images and inverse images of sets under functions?

REPLY [44 votes]: This big list is included in Appendix A of Introduction to Topological Manifolds by John M. Lee:
Let $f:X\to Y$ and $g:W\to X$ be maps, and suppose $R\subseteq W$, $S,S'\subseteq X$, and $T,T'\subseteq Y$.

$T\supseteq f(f^{-1}(T))$.
$T\subseteq T' \Rightarrow f^{-1}(T)\subseteq f^{-1}(T')$.
$f^{-1}(T\cup T')=f^{-1}(T)\cup f^{-1}(T')$.
$f^{-1}(T\cap T')=f^{-1}(T)\cap f^{-1}(T')$.
$f^{-1}(T\setminus T')=f^{-1}(T)\setminus f^{-1}(T')$.
$S\subseteq f^{-1}(f(S))$.
$S\subseteq S' \Rightarrow f(S)\subseteq f(S')$.
$f(S\cup S')=f(S)\cup f(S')$.
$f(S\cap S')\subseteq f(S)\cap f(S')$.
$f(S\setminus S')\supseteq f(S)\setminus f(S')$.
$f(S)\cap T=f(S\cap f^{-1}(T))$.
$f(S)\cup T\supseteq f(S\cup f^{-1}(T))$.
$S\cap f^{-1}(T)\subseteq f^{-1}(f(S)\cap T)$.
$S\cup f^{-1}(T)\subseteq f^{-1}(f(S)\cup T)$.
$(f\circ g)^{-1}(T)=g^{-1}(f^{-1}(T))$.
$(f\circ g)(R)=f(g(R))$.