Exercise 2.1

Proof. Consider any $x\in A_0$ and let $y=f(x)$ so that clearly $y\in f(A_0)$. Then, since $f(x)=y\in f(A_0)$, it follows from the definition of the preimage that $x\in f^{-1}(f(A_0))$. Hence $A_0\subset f^{-1}(f(A_0))$ as desired since $x$ was arbitrary. Now suppose that $f$ is also injective and consider this time any $x\in f^{-1}(f(A_0))$ so that $y=f(x)\in f(A_0)$ by the definition of a preimage. Then there is an $x^{\prime }\in A_0$ where $f(x^{\prime })=y=f(x)$ by the definition of an image. Since $f$ injective though, it must be that $x=x^{\prime }\in A_0$. This shows that $f^{-1}(f(A_0))\subset A_0$ since $x$ was arbitrary. The desired equality follows since it was already shown that $A_0\subset f^{-1}(f(A_0))$ (whether or not $f$ is injective). □

Proof. First suppose that $y$ is any element of $f(f^{-1}(B_0))$ so that there is an $x\in f^{-1}(B_0)$ where $f(x)=y$. Since $x\in f^{-1}(B_0)$, we then have that $y=f(x)\in B_0$ by the definition of a preimage. Hence $f(f^{-1}(B_0))\subset B_0$ since $y$ was arbitrary. Now suppose also that $f$ is surjective and suppose that $y\in B_0$ so that also clearly $y\in B$ since $B_0\subset B$. Since $f$ is surjective, there is an $x\in A$ where $f(x)=y$. We then have that $x\in f^{-1}(B_0)$ since $f(x)=y\in B_0$. Clearly then $y=f(x)\in f(f^{-1}(B_0))$ so that $B_0\subset f(f^{-1}(B_0))$ since $y$ was arbitrary. This shows equality as desired. □