Are Embeddings open functions?

Question

My book defines as topological embedding any continous injective function between topological spaces $i:X \mapsto Y$ such that $\forall E \subseteq X: E$ is open, $\exists F \subseteq Y : F$ is open and $E=i^{-1}(F)$. If that's the case, aren't all embeddings also open functions (that is they map open sets to open sets)? I thought so, in fact $E=i^{-1}(F) \implies i(E) = F$ which is, by definition, open. The problem is that then the book defines an "open embedding" as "an embedding that also happens to be an open function", so i guess that's not always the case. What am i doing wrong then?

Answer 1

No. All inclusions of points in a topological space are embeddings, but rarely open.

Your mistake iis that “$E = i^{-1}(F) \implies i(E) = F$” is just not true. That would imply, that the entirety of $F$ is hit by $i$. But setting $E = X$ and $F = Y$, certainly $E = i^{-1}(F)$, but $i(E) = F$ would mean that $i$ is surjective in this case.

In general, you only have “$E ⊆ i^{-1} F \iff i(E) ⊆ F$”, which is basically the definition of inverse images and a very important fact.

If a map $i$ is surjective, then indeed $i(i-^{1})(F) = F$ for all subsets $F ⊆ Y$, so the implication “$E = i^{-1}(F) \implies i(E) = F$” is true in this case. So surjective embeddings are open, hence homeomorphisms. That’s what you discovered.