If we have an embedding $f:X \rightarrow A$, where $A \subset Y$, do we have to show $f^{-1}$ is continuous?
I'm looking at a proof where they only show that $f$ is continuous and 1-1. Then I looked at the Wikipedia article on embeddings,
http://en.wikipedia.org/wiki/Embedding
"In general topology, an embedding is a one-to-one function (i.e., an injection) that is a homeomorphism onto its image.[1] More explicitly, an injective continuous map f : X → Y between topological spaces X and Y is a topological embedding if f yields a homeomorphism between X and f(X) (where f(X) carries the subspace topology inherited from Y). Intuitively then, the embedding f : X → Y lets us treat X as a subspace of Y. Every embedding is injective and continuous. Every map that is injective, continuous and either open or closed is an embedding; however there are also embeddings which are neither open nor closed. The latter happens if the image f(X) is neither an open set nor a closed set in Y."
I'm confused. Is there a difference, in topology, between an embedding and a topological embedding? Are there some cases where I don't have to show that $f^{-1}$ is continuous?
EDIT: $X,Y$ are compact metric spaces.
Best Answer
If, as in this case, $X$ is compact and $A$ is Hausdorff, it suffices to show that $f:X\to A$ is a continuous bijection. To see this, let $F$ be a closed set in $X$. Then $F$ is compact, and $f$ is continuous, so $f[F]$ is compact in $A$ and therefore closed in $A$, since compact sets in a Hausdorff space are closed. Thus, $f$ is a closed map. Now let $U$ be any open set in $X$; then $X\setminus U$ is closed, and $f[U]=A\setminus f[X\setminus U]$ is open, since $f$ is a bijection, and $f^{-1}$ is therefore continuous.