# Solution check of: $f:X \to Y$ surjective if and only if for any $T \subseteq Y,T \ne \varnothing$ it is $f^{-1}(T) \ne \varnothing$

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?

• What is your definition of surjective?
– cat
Commented Mar 22, 2022 at 1:22
• 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 Commented Mar 22, 2022 at 1:24