# Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$?

Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$?

I was trying to prove that there is no such function, but failed.

Any suggestions?

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It might be a good idea to post what you've tried :) – Sp3000 Jan 21 '12 at 14:25
one - one just means injective. There are many such maps. Only if you in addition require the map to be onto, then there is no such map. – user20266 Jan 21 '12 at 14:27
Thanks for the comment. – tomerga2 Jan 21 '12 at 14:43
Any 1-1 onto continous function from $[0,1]$ to $[0,1]^2$ would be a homeomorphism since we are dealing with compact spaces. (That is the inverse would automatically be continuous.) Can you think of a good reason the two spaces are not homeomorphic? What happens when you remove a point from each space? – Grumpy Parsnip Jan 21 '12 at 14:50
I really don't see how to solve this problem using only the methods of multivariable calculus. You learn almost nothing about continuous functions in this course...only about differentiable ones. On the other hand, a little general topology makes this problem quite easy: as others have said, the key is that since $[0,1]$ is compact, the hypotheses force $f$ to be a homeomorphism, and it is easy to see that the interval is not homeomorphic to the square. I wonder what your instructor has in mind here? (I doubt it's the implicit function theorem for nondifferentiable functions...) – Pete L. Clark Jan 21 '12 at 19:49

If such a function existed, it would have a continuous inverse. Because the preimage of a closed set under the inverse would just be the forward image of a closed set under the original function. Now every closed subset of $[0,1]$ is compact, so it will be mapped by a continuous function to a compact, and hence closed, subset of $[0,1]^2$.

So it suffices to show that no continuous function $f:[0,1]^2\to[0,1]$ is injective. For this, you can find two different paths in $[0,1]^2$ that only agree at the endpoints and use the intermediate value theorem to show that $f$ is not injective.

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Dear Michael: I googled "one-one" and found nothing. The closest match is one-to-one, which means "injective". --- I think the question you answered is (more interesting but) different from the one that was asked. (+1) --- Perhaps you answered the question the OP meant to ask (but not the one she/he asked). – Pierre-Yves Gaillard Jan 22 '12 at 5:08
From the comments after OPs post, he seems to agree with the interpretation of it meaning "injective". – Michael Greinecker Jan 22 '12 at 7:13
Dear Michael: Sorry, you're right. I missed the word "onto". – Pierre-Yves Gaillard Jan 22 '12 at 7:31

Here's an alternate solution using properties of connected sets:

Suppose for contradiction such a function $\varphi$ existed, then it would have a continuous inverse (cf. baby Rudin 4.17) $$\varphi^{-1} : [0,1]^2 \rightarrow [0,1]$$ Now, $[0,1]^2 \setminus\{\varphi(1/2)\}$ is connected (since $\varphi$ is a bijection we are just removing one point from the square) and by properties of continuous functions we know that $\varphi^{-1}$ maps connected sets to connected sets (cf. baby Rudin 4.22) . But $$\varphi^{-1}([0,1]^2 \setminus\{\varphi(1/2)\}) = [0, 1/2)\cup(1/2, 1]$$ which is not connected, so we have a contradiction and hence no such map can exist.

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Use the generalized Implicit Function Theorem. This is the sort of problem it should be useful for.

"Various forms of the implicit function theorem exist for the case when the function f is not differentiable..."

http://en.wikipedia.org/wiki/Implicit_function_theorem

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Continuous functions need not be differentiable. In fact there are continuous functions that are nowhere differentiable. – Zhen Lin Jan 21 '12 at 15:52
Then could this be useful? "Various forms of the implicit function theorem exist for the case when the function f is not differentiable..." (See Wikipedia: en.wikipedia.org/wiki/Implicit_function_theorem) – mathNotebook Jan 21 '12 at 16:00
This is an interesting idea, but it's hard for me to believe that a multivariable calculus student is supposed to know and use a theorem from the late 20th century in order to solve his homework problems. – Pete L. Clark Jan 21 '12 at 20:51