Show that $f:X \to Y$ surjective if and only if for any $T \subseteq Y,T \ne \varnothing$ it is $f^{-1}(T) \ne \varnothing$.

My work:

$\boxed{\Longrightarrow}$ Let $T\subseteq Y,T\ne\varnothing$. So there exists $y \in T$, and since by hypothesis $f:X \to Y$ is surjective there exists $x \in X$ such that $f(x)=y$. So, by definition of preimage, it is $x \in f^{-1}(T)$, hence $f^{-1}(T) \ne \varnothing$. Since $T\subseteq Y,T \ne \varnothing$ is arbitrary, this holds for any $T\subseteq Y,T \ne \varnothing$.

$\boxed{\Longleftarrow}$: Let $T\subseteq Y, T \ne \varnothing$ such that $f^{-1}(T) \ne \varnothing$. By hypothesis $T \ne \varnothing$ and $f^{-1}(T) \ne \varnothing$, so there exist $x \in X$ such that $f(x) \in T$. By hypothesis this holds for any $T \subseteq Y$, so in particular this holds for $\bigcup_{T \subseteq Y} T$; since $T\subseteq Y$, it is $Y=\bigcup_{T \subseteq Y} T$ hence for any $y \in Y$ there exist $x \in X$ such that $f(x)=y$, thus $f:X \to Y$ is surjective.

Is this proof correct?

  • $\begingroup$ What is your definition of surjective? $\endgroup$
    – cat
    Commented Mar 22, 2022 at 1:22
  • $\begingroup$ This looks pretty good. You could probably make the second direction a little cleaner though. Maybe try something like this: pick $y\in Y$. We need to find $x\in X$ with $f(x)=y$. Note that $\{y\}\subseteq Y$ is a nonempty subset of $Y$. Thus the assumption says that $f^{-1}(\{y\})$ is nonempty in $X$. So there is an element $x\in X$ such that $f(x)\in\{y\}$ and hence $f(x)=y$ as desired $\endgroup$
    – wormram
    Commented Mar 22, 2022 at 1:24


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