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Fix function name typos in exercise 2.7 injectivity proof #27

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hooyuser merged 2 commits into from
Dec 17, 2025
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Fix function name typos in exercise 2.7 injectivity proof #27

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Copilot Dec 17, 2025
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Fix typos in exercise 2.7: change f to g in injectivity proof
Copilot Dec 17, 2025
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6 changes: 3 additions & 3 deletions chapter1.tex
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Original file line number Diff line number Diff line change
Expand Up @@ -318,9 +318,9 @@ \subsection{\textsection2. Functions between sets}
Let's construct such a function $g$, defined to be $g(a,b) = a$. Keep in mind
that here $(a,b)\in\Gamma_f\subseteq A\times B$.

Let $(a',b'),(a'',b'')\in\Gamma_f$ such that $f(a',b') = f(a'',b'')$. For
contradiction, suppose that $(a',b')\neq (a'',b'')$. Since $f(a',b') = a' = a''
= f(a'',b'')$, it must be that $b'\neq b''$. However, this would mean that both
Let $(a',b'),(a'',b'')\in\Gamma_f$ such that $g(a',b') = g(a'',b'')$. For
contradiction, suppose that $(a',b')\neq (a'',b'')$. Since $g(a',b') = a' = a''
= g(a'',b'')$, it must be that $b'\neq b''$. However, this would mean that both
$(a',b')$ and $(a',b'')$ are in $\Gamma_f$; this would mean that $f(a') = b'
\neq b'' = f(a')$, which is impossible since $f$ is a function. Hence $g$ is
injective.
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