Fix typo in Chapter 1 Exercise 2.2: change f(g(a))=a to f(g(b))=b

Co-authored-by: hooyuser <[email protected]>

Author: Copilot

chapter1.tex

@@ -200,7 +200,7 @@ \subsection{\textsection2. Functions between sets}
 	exists an $a = g(b)$ such that $f(a) = b$. Hence $f$ is surjective.
 
 	($\impliedby$) Suppose that $f$ is surjective. We want to construct a function
-	$g:B\to A$ such that $f(g(a)) = a$ for all $a\in A$. Since $f$ is surjective,
+	$g:B\to A$ such that $f(g(b)) = b$ for all $b\in B$. Since $f$ is surjective,
 	for all $b\in B$ there is an $a\in A$ such that $f(a) = b$. For each $b\in B$
 	construct a set $\Lambda_b$ of such pairs:
 	%