Merge pull request #18 from hooyuser/copilot/fix-chapter-1-exercise-2-2

hooyuser · web-flow · commit d8e990ea82e7 · 2025-10-19T08:59:05.000+08:00
Fix typo in Chapter 1 Exercise 2.2 proof of Proposition 2.1
diff --git a/chapter1.tex b/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:
 	%

Original file line number	Diff line number	Diff line change
`@@ -200,7 +200,7 @@ \subsection{\textsection2. Functions between sets}`
`200`	`200`	`exists an $a = g(b)$ such that $f(a) = b$. Hence $f$ is surjective.`
`201`	`201`
`202`	`202`	`($\impliedby$) Suppose that $f$ is surjective. We want to construct a function`
`203`		`- $g:B\to A$ such that $f(g(a)) = a$ for all $a\in A$. Since $f$ is surjective,`
	`203`	`+ $g:B\to A$ such that $f(g(b)) = b$ for all $b\in B$. Since $f$ is surjective,`
`204`	`204`	`for all $b\in B$ there is an $a\in A$ such that $f(a) = b$. For each $b\in B$`
`205`	`205`	`construct a set $\Lambda_b$ of such pairs:`
`206`	`206`	`%`