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Invariance of domain at least in statement seems a simple result. I mean, the first time I saw the statement I thought: "the proof can't be that bad", but when I searched for it I saw that it needs even algebraic topology to prove this result.

My doubt is: isn't there any other proof of invariance of domain that don't need to use algebraic topology? Is there some more elementary proof of this result?

Thanks very much in advance!

Gold
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    If you're willing to stipulate differentiability, it's not nearly so difficult, but in the continuous world there are things like space-filling curves and horned spheres :) – Ted Shifrin Aug 25 '13 at 17:07
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    Yes that's true, with differentiability assumed there's one exercise of Spivak's Calculus on Manifolds to prove this result. I see that the problem is with just continuity. – Gold Aug 25 '13 at 17:09
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    One can ask the same question about other fundamental topological theorems, like Brouwer's fixed-point theorem, Jordan separation theorem and so on. As far as I know, the answer is almost invariably either "no" or "yes, but a proof is long and complicated as you have to rediscover some basic algebraic topology in disguise". – Moishe Kohan Aug 25 '13 at 19:09
  • Does this answer your question? A valid proof for the invariance of domain theorem? – Chain Markov Dec 24 '19 at 22:29
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    @TedShifrin Do you know such an easier proof for the smooth case? I could not find one, after a long search. You may see this question-https://math.stackexchange.com/questions/600027/domain-invariance-for-smooth-functions – Asaf Shachar May 21 '20 at 06:36
  • @user1620696 Do you know which exercise is it? Or what Spivak had in mind as a possible elementary proof? (You may see my comment above). – Asaf Shachar May 21 '20 at 06:40
  • @AsafShachar: Spivak assumes that the map has invertible derivative everywhere (Problem 2-36 in CoM). This makes the problem trivial indeed. – Moishe Kohan May 22 '20 at 15:46
  • @MoisheKohan Thanks, you are right. So, we are still left with the question whether there exist a simpler "elementary" proof of the $C^1$ case, without assuming invertibility. – Asaf Shachar May 22 '20 at 16:02
  • @MoisheKohan If I am not mistaken you have commented that there exist a proof under the assumption of $C^1$ of the inverse map $f^{-1}$ (using a fixed point argument)? Did you delete it? I couldn't find it now. – Asaf Shachar May 27 '20 at 08:36
  • @AsafShachar: I realized that simplifications are only marginal. While one does not need algebraic topology apparatus, one still has to use some transversality arguments. So, in the end, it is not that much shorter than Tao’s argument (combined with fixed point theorem). – Moishe Kohan May 27 '20 at 15:35

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The point here is what means elementary. Many think that a bit of singular homology is not that bad (see Greenberg-Harper's Algebraic Topology textbook, for instance). One can use Mapping Degree Theory as in many books devoted to the topic and use Borsuk-Ulam or else. But I would recommend a carefull reading of

http://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/

There one finds a very nice proof. The heavy part is Brouwer Fixed Point Theorem. Apart from that, some manipulations involving Weierstrass Approximation Theorem and the so-called Little Sard Theorem (in this particular case: a polynomial image of $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$ has empty interior, this disguised in measure zero terms).

Jesus RS
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