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I was working on problem 2-37(a) of Spivak's Calculus on Manifolds and I found that my approach to solving it was probably quite different from what was intended.

To be clear, I understand the argument Spivak probably intended the reader to make. Also, just to clarify, I'm not saying my approach is better. Indeed I think the intended approach is preferable as it utilizes the theorems of the section and anticipates some of the ideas used to prove the Implicit Function Theorem in the section ahead. I just wanted to see if there was something wrong with my approach, as it doesn't require the same assumptions that were given in the problem.

We are asked to show that if $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ is continuously differentiable then $f$ is not injective. We argue that, assuming $f$ is continuously differentiable (though we could use a weaker assumption than this) and injective, it follows that $f(x,y)$ is differentiable on $\mathbb{R}^2$ and therefore continuous on $\mathbb{R}^2$. So $f(x,0)$ and $f(0,y)$ are continuous and injective on the interval $[-1,1]\subset \mathbb{R}$. By connectedness of $[-1,1]$ and continuity of $f(x,0)$ and $f(0,y)$, the images of $[-1,1]$ by $f(x,0)$ and $f(0,y)$ are both connected. $f(0,0)$ being a maximum or minimum of either function on $[-1,1]$ contradicts with $f$ being injective. If $f(0,0)$ is not a maximum or a minimum, then it follows by connectedness that both images contain some open interval about $f(0,0)$, which, again, contradicts with injectivity of $f$.

Thanks in advance.

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    $\begingroup$ Why would you not want to go into details? $\endgroup$ Dec 16, 2017 at 18:25
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    $\begingroup$ You actually are using more than you realize. It takes a little work to prove that a one-to-one continuous function from an open interval of $\Bbb R$ to $\Bbb R$ is an open map. But this seems fine to me. It's far easier if you know it's differentiable with nonzero derivative on the interval. $\endgroup$ Dec 16, 2017 at 18:26
  • $\begingroup$ I just assumed if one of my conclusions was incorrect, it would be easier to point it out without having to find it in the details. I'll make an edit and flesh it out though. $\endgroup$
    – Tom H.
    Dec 16, 2017 at 18:29
  • $\begingroup$ @TedShifrin Oh my, I'm a little star struck. You're Professor Theodore Shifrin. I've actually been using your videos on youtube to supplement my self-study of Spivak's book. Your lectures are incredible. I'm actually saving up to buy your book to supplement my self-study as well. Thanks for taking the time to respond and for being a world-class teacher. $\endgroup$
    – Tom H.
    Dec 16, 2017 at 23:32
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    $\begingroup$ Thanks for the kind words, @CryinShame. You can often find me in chat if there are concepts you want to talk about more interactively. $\endgroup$ Dec 17, 2017 at 0:56

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Your proof is correct!


It is also worth pointing out that your proof can be generalised to show that there is no continuous injective function $f : \mathbb{R}^n \to \mathbb{R}$ for any $n > 1$.

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