Prove a mapping is open

Question

Prove that the map $$f:(0,1)\to\mathbb{R}^2$$ $$t\mapsto (\cos 2\pi t,\sin 2\pi t)$$ is an embedding.

PS: In general topology, an embedding is a homeomorphism onto its image. More explicitly, an injective continuous map $f:X\to Y$ between topological spaces $X, Y$ is a topological embedding if $f$ yields a homeomorphism between $X$ and $f(X)$.

I already prove $f$ is injective, continuous, but stuck at proving it yields a homeomorphism between $X$ and $f(X)$. To prove this, I'm trying to prove $g:X\to f(X)$ is homeomorphism $\iff g$ is an open mapping, which means $g(U)$ is open for every open set $U\subset X.$
Could someone help me to finish the problem? Thank in advance!

Answer 1

Let $g:X\rightarrow Y$ be an injective continuous map, then $g:X\rightarrow g(X)$ is continuous bijection. Equip $g(X)\subset Y$ with the subspace topology, and suppose that $g$ is a homeomorphism. Then it follows that there is a continuous inverse $g^{-1}:g(X)\rightarrow X$. Now let $U\subset X$ be open, then we see that by definition $(g^{-1})^{-1}(U)$ is an open subset of $g(X)$, but clearly $(g^{-1})^{-1}(U)=g(U)$, so $g(U)$ and $g$ is an open map.

Now suppose that $g:X\rightarrow g(X)$ is a continuous open bijection. By the fact that $g$ is a bijection, there exists a priori a set map $g^{-1}:g(X)\rightarrow X$. Our job is to show this is continuous, so let $U\subset X$ be open, then again we have that $(g^{-1})^{-1}(U)=g(U)$, but $g$ is open so $(g^{-1})^{-1}(U)$ is open, and $g^{-1}$ is continuous implying that $g$ is a homeomorphism.

Anyways, to show this in your specific case, it suffices to prove that $f$ is open on basic opens (i.e. on the basis open sets which generate the subspace topology) of $(0,1)$, that is intervals of the. form $(a,b)$ where $0<a, b<1$. We see that $f((a,b))$ is the arc of radius one which with end points at $a2\pi$ and $b2\pi$ radians. In other words, in polar coordinates on $\mathbb R^2$, we have that: $$f((a,b))=\{(r,\theta)\in \mathbb R^2: r=1, a2\pi<\theta<b2\pi\}$$ We also have that: $$f((0,1))=\{(r,\theta)\in\mathbb R^2:r=1, 0<\theta<2\pi\}$$ Let $V\subset \mathbb R^2$ be the open set: $$V=\{(r,\theta)\in \mathbb R^2:1-\epsilon <r<1+\epsilon, a2\pi<\theta<b2\pi\}$$ for some $\epsilon>0$. Then clearly we have that: $$f((a,b))=V\cap f((0,1))$$ so $f((a,b))$ is open in the subspace topology as desired.

Answer 2

It's enough to show that the images under $f$ of all elements of a subbasis for the domain $X = (0, 1)$ are open. Taking $X$ as a subspace of $\mathbb{R}$ under the order (usual) topology, one subbasis for $X$ is the collection of intervals $\{ (0, a) | 0 < a < 1 \} \cup \{ (b, 1) | 0 < b < 1 \}$.

The image of the interval $(0, a)$ ($0 < a < 1$) is an arc of the unit circle that does not include its endpoints $(1, 0)$ and $(\cos (2 \pi a), \sin (2 \pi a))$. This set is open in $f (X)$ as a subspace of $\mathbb{R}^2$ as a metric space with the Euclidean metric $d$: if you take any point $p = (\cos (2 \pi t), \sin (2 \pi t))$ in this set, it is contained in the set $f (X) \cap B_d \left( p, \sin \left( \pi \min \{ t, a - t \} \right) \right)$, which is open in $f (X) \subseteq \mathbb{R}^2$ and contained in the arc (as you should check).

You can perform a very similar analysis for images of subbasis elements of the form $(b, 1)$ ($0 < b < 1$).

Answer 3

To prove that the map $f: (0,1) \to \mathbb{R}^2$ defined by $t \mapsto (\cos 2\pi t, \sin 2\pi t)$ is an embedding, we need to show that it is an injective continuous map which is a homeomorphism onto its image. You've already proved that $f$ is injective.

It's evident that both components of $f$, namely $\cos 2\pi t$ and $\sin 2\pi t$, are continuous functions. Since $\cos$ and $\sin$ are continuous functions and composition of continuous functions is continuous, $f$ is continuous.

Now, to show that $f$ is a homeomorphism onto its image, we need to prove two things:

Onto its image: For any point $(x,y)$ in the unit circle $\mathbb{S}^1$, there exists $t$ such that $x = \cos 2\pi t$ and $y = \sin 2\pi t$ (since $\cos^2 t + \sin^2 t = 1$). Thus, every point in $\mathbb{S}^1$ is covered by $f$. Since $f$ is defined on $(0,1)$, it covers $\mathbb{S}^1$ except possibly at the points $(1,0)$ and $(-1,0)$. However, since $\cos$ and $\sin$ are periodic functions, the map $f$ covers $\mathbb{S}^1$ infinitely many times.

Homeomorphism: We need to show that $f$ is a continuous bijection and its inverse is continuous. To do this, it suffices to show that $f$ is a closed map (since $f$ is injective, it automatically has a continuous inverse, namely its restriction to its image).

To prove that $f$ is a closed map, let $A$ be any closed subset of $(0,1)$. Since $(0,1)$ is a closed subset of itself, $A$ is also closed in $(0,1)$. We want to show that $f(A)$ is closed in $\mathbb{R}^2$.

Consider the sequence $\{f(x_n)\}$ in $f(A)$ converging to some point $y$ in $\mathbb{R}^2$. We need to show that $y \in f(A)$. Since $f(x_n)$ converges to $y$, the sequences $\{\cos 2\pi x_n\}$ and $\{\sin 2\pi x_n\}$ must converge to $y_1$ and $y_2$ respectively, where $y = (y_1, y_2)$. Since $A$ is closed, the limit $x = \lim x_n$ is in $A$. Since $f(x_n) = (\cos 2\pi x_n, \sin 2\pi x_n)$ and both $\cos$ and $\sin$ are continuous functions, we have $f(x) = (\cos 2\pi x, \sin 2\pi x) = (y_1, y_2) = y$. Hence, $y \in f(A)$.

Thus, $f(A)$ is closed in $\mathbb{R}^2$, and $f$ is a closed map. Since it's already continuous and injective, it is indeed an embedding.

Answer 4

It is easier to show that $f : (0,1) \to f((0,1)) = S := S^1 \setminus \{ e_1 \}$, where $e_1 = (1,0) \in S^1$, is a closed map.

Consider the continuous map $$F : [0,1] \to S^1, F(t) = (\cos 2\pi t, \sin 2 \pi t).$$ Your map $f : (0,1) \to S$ is a restriction of $F$.

Let $C \subset (0,1)$ be closed in $(0,1)$ and $C' = \overline{C}^{[0,1]}$ be its closure in $[0,1]$. Then $C'$ is compact, hence also $F(C')$ is compact, thus closed in $S^1$. Therefore $F(C') \cap S$ is closed in $S$.

We have $C' \cap (0,1) = \overline{C}^{[0,1]} \cap (0,1) = \overline{C}^{(0,1)} = C$. With $Z = C' \cap \{0,1\} \subset \{0,1\}$ we obtain

$$C' = C' \cap (0,1) \cup C' \cap \{0,1\} = C \cup Z.$$

Since $F(Z) \subset \{ e_1 \}$ we get $$F(C') \cap S = F(C \cup Z) \cap S = (F(C) \cup F(Z)) \cap S = (f(C) \cup F(Z)) \cap S \\= f(C) \cap S \cup F(Z) \cap S = f(C) .$$ This proves that $f(C)$ is closed in $S$.