Definition: Let $X$ and $Y$ be topological spaces. Suppose that $Y$ is compact and Hausdorff. Let $f,g:X\to Y$ be embeddings. We say that $f$ is ambient isotopic to $g$ (denote $f\sim g$) if there is a continuous function $F:Y\times [0,1]\to Y$ such that the functions of the form $F_t:Y\to Y$, $F_t(y)=F(y,t)$ for $t\in[0,1]$, satisfy the following:
- $F_t$ is a homeomorphism for all $t\in[0,1]$.
- $F_0=id_Y$ and $F_1 \circ f=g$.
The map $F$ is called the ambient isotopy from $f$ to $g$.
Question: If $f\sim g$, is true that $g\sim f$ ?
I think that the answer is yes, and this is what I have done:
Let $F$ be the ambient isotopy from $f$ to $g$. Consider $G:Y\times [0,1]\to Y$, $G(y,t):=F_t^{-1}(y)$. It is clear that $G$ satisfies conditions 1 and 2 of the definition. How can I prove that $G$ is continuous? I don't know. But here we have some facts:
- Every closed subset in $Y$ is compact.
- If $U\subset Y$, then $G^{-1}(U)=\bigcup_{t\in [0,1]}F_t(U)\times\{t\}$.
- $F(U\times [0,1])=\bigcup_{t\in [0,1]}F_t(U)$, therefore $\pi_1(G^{-1}(U))=F(U\times [0,1])$, where $\pi_1$ is the projection over the first component.
May you give me a hand please?
My previous answer was incorrect, as @Chilote pointed out. Here is the correct answer: $$G(y,t) = F_1^{-1}(F(y,1-t)) $$ This is continuous and each $G_t(y) = G(t,y)$ is a homeomorphism. Also, $$G_0(y) = G(y,0) = F_1^{-1}(F(y,1))=F_1^{-1}(F_1^{\vphantom{-1}}(y))=y $$ Also, based on the equation $f = F_1^{-1}\circ g$ we have $$G_1(g(y)) = G(g(y),1) = F_1^{-1}(F(g(y),0)) = F_1^{-1}(g(y)) = f(y) $$ So to summarize, you don't have to work so hard to "reprove" continuity, you simply use appropriate compositions and inversions so that continuity is obvious.