Ambient Isotopic Embeddings on Topological Spaces
Recall from the Isotopic Embeddings on Topological Spaces page that if $X$ and $Y$ are topological spaces and $f, g : X \to Y$ are embeddings, then $f$ is said to be isotopic to $g$ if there exists a continuous function $H : X \times I \to Y$ such that $H_t : X \to Y$ is an embedding for every $t \in I$, $H_0 = f$, and $H_1 = g$.
We also proved that isotopy is an equivalence relation on the set of embeddings from $X$ to $Y$.
We will now describe another type of isotopy.
Definition: Let $X$ and $Y$ be topological spaces and let $f, g : X \to Y$ be embeddings. Then $f$ is Ambient Isotopic to $g$ in $Y$ if there exists a continuous function $H : Y \times I \to Y$ such that: 1) $H_t : Y \to Y$ is a homeomorphism for every $t \in I$. 2) $H_0 = \mathrm{id}_Y$. 3) $H_1 \circ f = g$. If such a function $H$ exists, then $H$ is said to be an Ambient Isotopy from $f$ to $g$. |
Definition: Let $A$ and $B$ be subspaces of a topological space $Y$. Then $A$ and $B$ are said to be Ambient Isotopic within $Y$ if there exists a continuous function $H : Y \times I \to Y$ such that: 1) $H_t : Y \to Y$ is a homeomorphism for every $t \in I$. 2) $H_0 = \mathrm{id}_Y$. 3) $H_1(A) = B$. |
For example, consider the spaces $X = \{ a \}$ and $Y = [0, 1]$. Let $f : \{ a \} \to [0, 1]$ be defined by $f(c) = \frac{1}{3}$ and let $g : \{ a \} \to [0, 1]$ be defined by $g(a) = \frac{2}{3}$. We will show that $f$ is ambient isotopic to $g$ within $[0, 1]$.
Define a function $H : [0, 1] \times [0, 1] \to [0, 1]$ by:
(1)Then $H$ is continuous by the gluing lemma. Furthermore, $H_0(y) = H(y, 0) = y = \mathrm{id}_Y$ and $H_1(f(c)) = H(f(c), 1) = 1 - f(c) = g(c)$, so $H_1 \circ f = g$.
Theorem 1: Let $X$ and $Y$ be topological spaces and let $f, g : X \to Y$ be embeddings. If $f$ is ambient isotopic to $g$ in $Y$ then $f$ is isotopic to $g$. |
- Proof: Since $f$ is ambient isotopic to $g$ in $Y$ there exists a continuous function $H : Y \times I \to Y$ such that $H_t$ is a homeomorphism for every $t \in I$, $H_0 = \mathrm{id}_Y$, and $H_1 \circ f = g$. Consider the function $K : X \times I \to Y$ defined by:
- Then $K$ is a continuous function since $f$ and $H$ are continuous functions. Furthermore, $K_t = H_t \circ f$ is an embedding for every $t \in I$. Lastly, $K_0(x) = H(f(x), 0) = H_0(f(x)) = f(x)$, i.e., $K_0 = f$, and $K_1(x) = H(f(x), 1) = H_1 (f(x)) = g(x)$, so $K_1 = g$. Therefore $f$ is isotopic to $g$. $\blacksquare$