Topology, 2ed

Let $f:A\to B, A_0\subseteq A, B_0\subseteq B$.
(a) Show that $A_0\subseteq f^{-1}(f(A_0))$ and that equality holds if $f$ is injective.
(b) Show that $f(f^{-1}(B_0))\subseteq B_0$ and that equality holds if $f$ is surjective.

(a) Let $a\in A_0$. $a=f^{-1}(f(a))\in f^{-1}(f(A_0))\implies A_0\subseteq f^{-1}(f(A_0))$. Now suppose $f$ is injective, and let $a\in f^{-1}(f(A_0))$. $\exists b\in f(A_0)$ s.t. $a=f^{-1}(b)$. Since $b\in f(A_0),\exists a^\prime\in A_0$ s.t. $f(a^\prime)=b$. Since $f$ is injective, we must have $a=a^\prime$; that is, $a\in A_0$. Hence $f^{-1}(f(A_0))=A_0$.

(b)

Let $f:A\to B$, and let $A_i\subseteq A$ and $B_i\subseteq B$ for $i\in\{0,1\}$. Show that:
(a) $B_0\subseteq B_1 \implies f^{-1}(B_0)\subseteq f^{-1}(B_1)$
(b) $f^{-1}(B_0\cup B_1) = f^{-1}(B_0)\cup f^{-1}(B_1)$
© $f^{-1}(B_0\cap B_1) = f^{-1}(B_0)\cap f^{-1}(B_1)$
(d) $f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1)$
(e) $A_0\subseteq A_1\implies f(A_0)\subseteq f(A_1)$
(f) $f(A_0\cup A_1)=f(A_0)\cup f(A_1)$
(g) $f(A_0\cap A_1)\subseteq f(A_0)\cap f(A_1)$; equality holds if $f$ is injective
(h) $f(A_0 - A_1)\subseteq f(A_0) - f(A_1)$; equality holds if $f$ is injective