diff --git a/ch1/solutions.tex b/ch1/solutions.tex
index 38a5ced..05a1c5a 100644
--- a/ch1/solutions.tex
+++ b/ch1/solutions.tex
@@ -436,9 +436,9 @@ \subsection{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.