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Does there exist a continuous onto map from (-1,1] to (-1,1)? My thoughts. I think there doesn't exist any continous onto map satisfying this.

Proof: Suppose there exist such $f:(-1,1]$ to $(-1,1)$. Since if we take [a,1] where a is between 1<a<1, [a,1] is compact in (-1,1] Because [a,1] is compact in $R$, so $f[a,1]$ will be compact too, say it's image will be compact too. But atleast (-1,b) and (c,1) will never be in image of [a,1] for that some b,c. But now by my intuition, I think that there exist g such that -1<g<a and image of f(-1,g) will be connected so we will left with image of f(-1,g)=(-1,h)U(j,1). That will be the contradiction. But not able to think in regrous way I am trying to prove using this kind of argument. Any help will be appreciated

Gary
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Tony
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1 Answers1

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Such maps do in fact exist. For example, the function $(1-x)\sin\frac{1}{x}$ is continuous on $(0,1]$, with range $(-1,1)$; you can easily transform this into a function continuous on $(-1,1]$.

TonyK
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  • Thankyou sir. I have found the map as you suggest $f(x)=(\frac{1-x}{2})sin\frac{2}{x+1}$ – Tony May 06 '22 at 12:32