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Let $\beta(t) = (\sin 2t, \sin t) :(-\pi, \pi) \to \mathbb{R}^2$. This is an injective immersion and so its image $S$ is an immersed submanifold of $\mathbb{R}^2$ (the figure-eight). Let $G(t) = (\sin 2t, \sin t) : \mathbb{R} \to S$ where $S$ is with the topology and smooth structure as the immersed submanifold. The formula is the same with $\beta$, but the domain and codomain are different from those of $\beta$.

Example 5.28 of "Introduction to smooth manifolds" by J. Lee states that the map $G$ is not continuous because $\beta^{-1} \circ G$ is not continuous at $t = \pi$. In my understanding, $(\beta^{-1} \circ G)(\pi) = 0$ and the preimages of narrow intervals containing $0$ don't contain $\pi$, so it is not continuous there.

However, directly in terms of the open sets in $S$, I couldn't find any open set in $S$ whose preimage by $G$ is not open in $\mathbb{R}$. There must be such open sets in $S$ because $G$ is not continuous. What such open sets are there?

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cleone
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  • My copy of Lee's "Introduction to Smooth Manfiolds" does not have an Example 5.28. Presumably this example is only in the 2nd edition of the book, which came out in 2012. – Jim Belk Mar 09 '17 at 12:03
  • @JimBelk Google books shows the example in the second edition here – leeto Mar 09 '17 at 16:30

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The point is that the open subsets of $S$ with its given submanifold topology are not the same as the open sets in the subspace topology. For example, the subset $\beta\big( (-\pi/2,\pi/2)\big)$ (which looks like a backward S) is open in $S$, because we define the topology of $S$ in such a way that $\beta$ is a homeomorphism onto $S$. But the preimage of this set under $G$ contains $\pi$ as an isolated point.

[In response to @JimBelk's comment: Yes, the OP is referring to the second edition. The corresponding example in the first edition is Example 7.2.]

Jack Lee
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