From 0522d88cf207d984bb4e3aacf25df41ff4ca6c2c Mon Sep 17 00:00:00 2001
From: "copilot-swe-agent[bot]" <198982749+Copilot@users.noreply.github.com>
Date: Wed, 17 Dec 2025 10:30:12 +0000
Subject: [PATCH 1/2] Initial plan


From 38a1f7bad5ae1aa16be65d41f30a821d18c5c7f5 Mon Sep 17 00:00:00 2001
From: "copilot-swe-agent[bot]" <198982749+Copilot@users.noreply.github.com>
Date: Wed, 17 Dec 2025 10:32:34 +0000
Subject: [PATCH 2/2] Fix typos in exercise 2.7: change f to g in injectivity
 proof

Co-authored-by: hooyuser <10691820+hooyuser@users.noreply.github.com>
---
 chapter1.tex | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/chapter1.tex b/chapter1.tex
index b3153d9..613e3d2 100644
--- a/chapter1.tex
+++ b/chapter1.tex
@@ -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.