# Section 2.2: Problem 23 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Let $\mathfrak{A}$ be a structure and $g$ a one-to-one function with $\mbox{dom}g=|\mathfrak{A}|$ . Show that there is a unique structure $\mathfrak{B}$ such that $g$ is an isomorphism of $\mathfrak{A}$ onto $\mathfrak{B}$ .
This is similar to the previous exercise. Let , and we define the structure having the universe $|\mathfrak{B}|$ . Note that $g:|\mathfrak{A}|\rightarrow|\mathfrak{B}|$ and is bijective. For any $b\in|\mathfrak{B}|$ , let $h(b)$ be the point such that $g(a)=b$ . For any predicate symbol $P$ , let $\in P^{\mathfrak{B}}$ iff $\in P^{\mathfrak{A}}$ . Similarly, for any function symbol $f$ , let $f^{\mathfrak{B}}(b_{1},\ldots,b_{n})=g(f^{\mathfrak{A}}(h(b_{1}),\ldots,h(b_{n})))$ . For any constant symbol $c$ , let $c^{\mathfrak{B}}=g(c^{\mathfrak{A}})$ . Then, for any predicate parameter, $\in P^{\mathfrak{A}}$ iff $\in P^{\mathfrak{A}}$ iff $\in P^{\mathfrak{B}}$ . For any constant parameter, $g(c^{\mathfrak{A}})=c^{\mathfrak{B}}$ . And for every function parameter, $g(f^{\mathfrak{A}}(a_{1},\ldots,a_{n}))=g(f^{\mathfrak{A}}(h(g(a_{1})),\ldots,h(g(a_{n}))))=f^{\mathfrak{B}}(g(a_{1}),\ldots,g(a_{n}))$ . Therefore, $g$ is an isomorphism of $\mathfrak{A}$ onto $\mathfrak{B}$ .