I would like to prove following
Proposition 1. Define arbitrary $f: X \rightarrow Y$. If exists $U \subseteq Y$ such that $f\big(f^{-1}(U)\big) ≠ U$, then $f$ is not surjective.
Notes on notation:
- $f(U)$ denotes forward image, i.e
$$f(U) = \big\{f(x) \mid x \in U\big\}$$
- $f^{-1}(U)$ denotes backward image, i.e
$$f^{-1}(U) = \big\{ x \in \text{Dom}(f) \mid f(x) \in U\big\}$$
My attempt:
Lemma 1. For arbitrary $f: X \rightarrow Y$ and arbitrary $U \subseteq Y$, $f\big(f^{-1}(U)\big) \subseteq U$
Take some $y \in f\big(f^{-1}(U)\big)$. Then exists $x \in f^{-1}(U)$ such that $f(x) = y$. Since $x \in f^{-1}(U)$, $f(x) \in U$, and thus $y \in U$. $\Box$
Coming back to proposition 1:
Suppose there exists $U \subseteq Y$ such that $f(f^{-1}(U)) ≠ U$. By lemma 1, we know that $$f\big(f^{-1}(U)\big) \subseteq U\,.$$ Hence, it must be the case that $$U \not\subseteq f\big(f^{-1}(U)\big)\,.$$ Then we can take some $y \in U$, such that there is no $x \in f^{-1}(U)$ with $f(x) = y$. But that means that there is no $x \in X$ such that $f(x) = y$. And since $U \subseteq Y$, $y \in Y$, which means that $f$ is not surjective. $\Box$
Is it correct?
Are there any alternatives?