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I'm confused by the proof that $\epsilon$-$\delta$ continuity is equivalent to open-set continuity. One can prove that a function is $\epsilon$-$\delta$-continuous if and only if the preimage of any open set is open. However, it is not true that the image of any open set is open. My question is, what is the difference between image and preimage? The duality of image and preimage suggests to me that a proof that works for one of them should work for the other, but clearly this is not the case. Why not?

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    $\begingroup$ In the study of homeomorphism, you would see some functions, such $f:U\rightarrow W$, which is continuous, but $f^{-1}$ is not continuous. Therefore if, in your proof, there's no difference between image and preimage, then every continuous function must be a homeomorphism. $\endgroup$ Dec 19, 2014 at 17:39
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    $\begingroup$ I wouldn't say that there's any duality between preimage and image. For one thing, the preimage of the intersection is the intersection of the preimages, but the image of the intersection is not the intersection of the images. $\endgroup$
    – xyzzyz
    Dec 19, 2014 at 17:40
  • $\begingroup$ What exactly do you mean by 'duality of image and preimage'? The image of a constant function, which is continuous, is usually closed. $\endgroup$
    – copper.hat
    Dec 19, 2014 at 17:40

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The difference is in the following:

1) $x$ close to $y$ implies $f(x)$ is close to $f(y)$ : true for continuous functions

2) $f(x)$ close to $f(y)$ implies $x$ is close to $y$: generally false for continuous functions

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    $\begingroup$ hello ! did you intend to use x and y as sets ? I'm not sure what means your answer... I know that any pre image of a closed set is closed by a continuous function. Aren't you saying the opposite in your second sentence ? Kind regards. $\endgroup$ Mar 17, 2019 at 20:13
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    $\begingroup$ I'm not saying your answer is false, just that I don't understand it. If you don't want to answer, that's fine, but I can't ask someone else to explain what was on your mind, even more when yourself don't want to detail what was on your mind. I think your answer was really deep and interesting, if anyone coming by can explain me what @Umberto P. had on his mind is welcomed. $\endgroup$ Mar 17, 2019 at 20:41
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Any conjectured "symmetry" between preimage and image will fail because functions are not necessarily one-to-one. Notice that in the case of $f(x) = x^2$, the image of $(-1,1)$ is $[0,1)$, which is not open. The reason this fails is because the map has no inverse. If we reflect the graph across the line $y=x$, we end up with something that is not a function (it's not single valued). The preimage of $(-1,1)$ under this relation is $[0,1)$ but the result is not a continuous function because it is not a function.

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  • $\begingroup$ By "one-to-one" did you mean "a bijection"? Because "one-to-one" doesn't imply "onto" ("one-from-one"), but it seems like that's what you intended. $\endgroup$
    – user541686
    Dec 20, 2014 at 2:29
  • $\begingroup$ @Mehrdad By "one-to-one" I simply mean injection. I'm talking about the asymmetry in that functions are single valued ($x = y \implies f(x) = f(y)$) but not necessarily one-to-one ($f(x) = f(y) \implies x = y$). $\endgroup$
    – A.S
    Dec 20, 2014 at 6:21
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You can even have a continuous bijection that doesn't map open sets to open sets. Consider the function $$ f: [0,1) \cup [2,3] \to [0,2], \; f(x) = \begin{cases}x , & x \in [0,1) \\ x-1, & x \in [2,3] \end{cases}. $$

This is a continuous bijection. The set $[2,3]$ is open in $[0,1) \cup [2,3]$. But the image of $[2,3]$ is $[1,2]$ which is not open in $[0,2]$.

What happens with this function is that the domain is not connected, but the range is. So the function "glues" the two parts of the domain together continuously. This means that the inverse function has to rip apart these parts, so it's not continuous.

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