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Let $$f_1:U_1 \to \mathbb{R}^2 \quad f_1(t)=(\cos t, \sin t)\quad U_1=(-\pi, \pi)$$ $$f_2:U_2 \to \mathbb{R}^2 \quad f_2(t)=(\cos t, \sin t)\quad U_2=(\pi/2, 3\pi/2)$$

Show that $f_1^{-1}\bigg(f_1(U_1) \cap f_2(U_2)\bigg) $ is open.


I tried various methods but I keep running into problems.

For example the intersection is an arc, which is not open in $\mathbb{R}^2$, so I cannot just claim the statement is true due to continuity of $f$.

So I tried in the most basic way:

$p\in f_1^{-1}\bigg(f_1(U_1) \cap f_2(U_2)\bigg)\subset f^{-1}(f(U_1)) \implies \exists r\in U_1$ such that $f_1(r)=f_1(p)$

Because $U_1$ is open, $r+\epsilon\in U_1$, but $f(r+\epsilon)$ is not necessarily in the intersection.

Not sure what else to do.


Using DeMorgans

$f_1^{-1}\bigg(f_1(U_1) \cap f_2(U_2)\bigg) = f_1^{-1}(f_1(U_1)) \cap f_1^{-1}(f_2(U_2)) \supset U_1 \cap U_2$

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  • $\begingroup$ Inverse images commute with intersections. $\endgroup$ – Randall Dec 14 '17 at 22:07
  • $\begingroup$ @Randall, Cannot belief I forgot about DeMorgan's I'll give it a shot $\endgroup$ – PMM Dec 14 '17 at 22:11
  • $\begingroup$ @Randall Actually I had already tried that approach. I don't see what to do with it because you have that the preimage of the image of a set is a supset of that set. $\endgroup$ – PMM Dec 14 '17 at 22:17
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A useful lemma:

Let $g : X \to Y$ be an injection and $A \subseteq X$. Then $g^{-1}(g(A)) = A$.

Taking the preimage commutes with $\cap$ so we have:

$$f_1^{-1}(f_1(U_1) \cap f_2(U_2)) = f_1^{-1}(f_1(U_1)) \cap f_1^{-1}(f_2(U_2))$$

$f_1$ is injective so $f_1^{-1}(f_1(U_1)) = U_1$.

On the other hand, $$f_2(U_2) = f_1\left(\left\langle -\pi, -\frac{3\pi}2\right\rangle \cup \left\langle \frac\pi2, \pi\right\rangle\right) \cup \{(-1,0)\}$$

but $(-1, 0) \notin \operatorname{Im} f_1$ so

$$f_1^{-1}(f_2(U_2)) = f_1^{-1}\left(f_1\left(\left\langle -\pi, -\frac{3\pi}2\right\rangle \cup \left\langle \frac\pi2, \pi\right\rangle\right)\right) = \left\langle -\pi, -\frac{3\pi}2\right\rangle \cup \left\langle \frac\pi2, \pi\right\rangle$$

again because $f_1$ is injective.

So

$$f_1^{-1}(f_1(U_1) \cap f_2(U_2)) = U_1 \cap \underbrace{\left(\left\langle -\pi, -\frac{3\pi}2\right\rangle \cup \left\langle \frac\pi2, \pi\right\rangle\right)}_{\subseteq U_1} = \left\langle -\pi, -\frac{3\pi}2\right\rangle \cup \left\langle \frac\pi2, \pi\right\rangle$$

which is certainly an open set in $U_1$.

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  • $\begingroup$ The lemma is what I was missing mostly. Regarding notation, are the angle brackets open or closed sets? $\endgroup$ – PMM Dec 14 '17 at 22:31
  • $\begingroup$ @PMM The angle brackets are open intervals: $$(x,y) = \langle x, y \rangle$$ Of course, you can calculate $f_1(U_1)$ and $f_2(U_2)$ explicitly as they are simply arcs on the unit circle. Then intersect them, and notice that the intersection is open and $f_1$ is continuous. $\endgroup$ – mechanodroid Dec 14 '17 at 22:39
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Best if you draw it in a picture.

The point $(\cos t,\,\sin t)$ is just the point on the unit circle the radius of which encloses angle $t$ with the positive ray of $x$ axis. This angle $t$ is measured in radian, i.e. $\pi$ means $180^\circ$.

So, $f_1$ goes a full circle, not containing the starting = endpoint = $(-1,0)$ where $t=\pm\pi$. In other words, $f_1(U_1)$ is the circle minus this point $(-1,0)$.

And $f_2$ goes the left half circle from $\pi/2$ i.e. $90^\circ$ to $3\pi/2$ i.e. $270^\circ$, again not containing the endpoints.

Looking at the picture, clearly $f_1(U_1)\cap f_2(U_2)$ is $f_2(U_2)\setminus\{(-1,0)\}$.
Writing back the points of this to angles, using the range $U_1=(-\pi,\pi)$ we get exactly its preimage along $f_1$, which will be a union of two open intervals.

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