Merge pull request #27 from hooyuser/copilot/fix-typos-in-exercise-2-7

Fix function name typos in exercise 2.7 injectivity proof

Author: hooyuser

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.