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This question already has an answer here:

Here is my question: Is a homeomorphism, that is, a continuos function whose inverse function is also continuous, always bijective?

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marked as duplicate by Eric Wofsey general-topology Apr 2 '18 at 5:47

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    $\begingroup$ A function that isn't bijective doesn't have a well-defined inverse. Only bijective functions have inverses. $\endgroup$ – MJD Jun 16 '14 at 2:59
  • $\begingroup$ @MJD Oh! My bad, I'd better delete this question, so stupid. $\endgroup$ – user136592 Jun 16 '14 at 3:00
  • $\begingroup$ Where do you get a definition for homeomorphism that is not "a continuous bijection whose inverse is continuous". I.e., whoever defines homeomorphism without explicitly requiring bijectivity? $\endgroup$ – Eric Towers Jun 16 '14 at 3:01
  • $\begingroup$ @MJD: Don't mean to nitpick, but you may have 1-sided inverses in some "categories" $\endgroup$ – user99680 Jun 16 '14 at 3:03
  • $\begingroup$ @user99680 You can have partial inverses of nonbijective functions also. For example, $f:x\mapsto \sqrt x$ is a partial inverse of $g:x\mapsto x^2$, since we have $g(f(x)) = x$ for all $x$ for which the left side is defined. But that is not what is normally meant by the inverse of a function, and it is not what I guessed that OP meant. $\endgroup$ – MJD Jun 16 '14 at 3:06
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Homeomorphisms are always onto. However there is a possible source for a confusion in certain circumstances. One example is when you have a topological embedding $f : X \rightarrow Y$. In such a case, if $f$ is not surjective then $f$ is not a homeomorphism onto $Y$, however $f$ is a homeomorphism onto its image.

You may, e.g., embed the circle $S^1$ in the plane $\mathbb R^2$ by, say $f(t)=(\cos t,\sin t)$. This map is not a homeomorphism onto the target space $\mathbb R^2$, (it cannot be a homeomorphism onto $Y$, since $f$ is not onto $Y$) but it is a homeomorphism onto its image, and, tautologically, it is onto its image. Still, note that this is not entire a trivial issue since, e.g., by some results, $S^1$ may not be embedded into $\mathbb R$ , and, in general, $S^n$ cannot be embedded into $\mathbb R^n$, nor $\mathbb R^k ; k=0,1,2,..,n-1$

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  • $\begingroup$ A homeomorphism onto its image is bijective, just not with the larger space but rather the image itself. Indeed, this is precisely what "onto its image" means. I believe that to say that such a thing is not bijective really confuses the issue. $\endgroup$ – RghtHndSd Jun 16 '14 at 3:10
  • $\begingroup$ @rghthndsd: true, but how does one then make a distinction between a homeomorphism and an embedding? I think it is not a crucial issue, but I think it is not totally-trivial either. $\endgroup$ – user99680 Jun 16 '14 at 3:11
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    $\begingroup$ I don't understand. An embedding is an injective map $f : X \rightarrow Y$ with certain additional properties. Then $f : X \rightarrow \mathrm{Img}(X)$ is a homeomorphism. The statement "a homeomorphism need not be bijective" is false - I see no way around this. $\endgroup$ – RghtHndSd Jun 16 '14 at 3:14
  • $\begingroup$ @rghthndsd Well, but the issue is onto what: onto the host space or onto its image; the fact that every map is onto its image is tautological. Still, the point I am trying to address is that, given spaces $X,Y$ , $X$ may not always be embeddable into $Y$. Can you suggest how to make this point better than I did? $\endgroup$ – user99680 Jun 16 '14 at 3:16
  • $\begingroup$ When you say "homeomorphism onto its image" then you are specifying precisely onto what: it's onto its image. Where is the ambiguity? If you give me a map $f : X \rightarrow Y$ that is not onto, then $f$ is not a homeomorphism. Thus the statement "You may have a special case of a non-bijective homeomorphism" is false. I don't understand where you are coming from with the last sentence: what does the embeddability of certain spaces have to do with the definition of a homeomorphism? $\endgroup$ – RghtHndSd Jun 16 '14 at 3:21

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