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,


"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.

  • $\begingroup$ In the proof at which you’re looking, what else is known about $X$? If, for instance, $X$ is compact, it’s enough to show that $f$ is continuous and $1$-$1$. $\endgroup$ Aug 23 '13 at 23:41
  • 1
    $\begingroup$ @BrianM.Scott: ... and $Y$ or $A$ are Hausdorff, ... $\endgroup$ Aug 23 '13 at 23:42
  • $\begingroup$ Yes, $X,Y$ are compact. I was thinking about that,...what does that do? $\endgroup$
    – user41728
    Aug 23 '13 at 23:42
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    $\begingroup$ Every continuous bijection from a compact space to a Hausdorff space is a homeomorphism (should be in any standard intro to topology course/textbook). here's a proof $\endgroup$
    – Dan Rust
    Aug 23 '13 at 23:44
  • $\begingroup$ @DanielRust: You are right, I found it. I didn't expect this. Thanks. $\endgroup$
    – user41728
    Aug 23 '13 at 23:47

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.

  • $\begingroup$ Why do you conclude from this that $f^{-1}$ is continuous? $\endgroup$ Nov 27 '18 at 1:41
  • $\begingroup$ Because $f$ is an open map and $(f^{-1})^{-1} = f$. $\endgroup$
    – Moritz
    Oct 3 '19 at 11:57

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