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Fix typo in Chapter 1 Exercise 2.2 proof of Proposition 2.1 #18

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hooyuser merged 2 commits into from
Oct 19, 2025
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Fix typo in Chapter 1 Exercise 2.2 proof of Proposition 2.1 #18

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Copilot Oct 19, 2025
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Fix typo in Chapter 1 Exercise 2.2: change f(g(a))=a to f(g(b))=b
Copilot Oct 19, 2025
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2 changes: 1 addition & 1 deletion chapter1.tex
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Expand Up @@ -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:
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