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| -rw-r--r-- | library/function.tex | 7 |
1 files changed, 4 insertions, 3 deletions
diff --git a/library/function.tex b/library/function.tex index c7f709e..2585816 100644 --- a/library/function.tex +++ b/library/function.tex @@ -490,7 +490,7 @@ \begin{proposition}\label{inj_circ} Let $f\in\inj{A}{B}$. Let $g\in\inj{B}{C}$. - Then $g\circ f\in\surj{A}{C}$. + Then $g\circ f\in\inj{A}{C}$. \end{proposition} \begin{proof} Omitted. % TODO @@ -554,6 +554,7 @@ there exists $a\in\dom{f}$ such that $f(a) = b$ by \cref{fun_ran_iff}. Fix $b\in B$. Take $a\in A$ such that $f(a) = b$ by \cref{surj}. + Now $A=\dom{f}$. \end{proof} \begin{corollary}\label{funs_surj_iff} @@ -618,8 +619,8 @@ \end{proposition} \begin{proof} $g\circ f\in\surj{A}{C}$ by \cref{bijections,surjections_circ}. - $g\circ f$ is an injection. - Follows by \cref{circ_rightunique,bijections}. + $g\circ f$ is injective by \cref{bijections,circ_injective}. + Follows by \cref{bijections}. \end{proof} \subsection{Converse as a function} |