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6 changes: 3 additions & 3 deletions ch1/solutions.tex
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Expand Up @@ -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.
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