I am struggling to prove that every knot is isotopic to the unknot. Can someone please point me towards a reference of a proper proof? My attempt is as follows.
$\newcommand{\rthree}{\mathbf{R}^3}$ Let $K \subset \rthree$ be a knot.
Let $p \in K$. Take a ball $U_p \subset \rthree$ of radius $\varepsilon$ around the point $p$ such that the pair $(U_p, U_p \cap K)$ is homeomorphic to $(B, d)$, where $B$ is the unit ball in $\rthree$ centered at origin and $d$ is the diameter of $B$ along the $x$-axis. Such a ball exists as each point of the knot is locally flat. We choose a parametrization $f: [0, 2 \pi) \rightarrow K$ of the knot such that $f ([a, b]) = K \setminus (U_p \cap K)$, where $[a, b] \subset [0, 2 \pi)$.
Let $r \in \rthree$ be point outside $U_p$, i.e. $r \notin U_p$. Now consider the function $i_t \colon K \rightarrow \rthree$ defined for each $t \in [0, 1]$ as follows. $$ i_t (f(x)) = \begin{cases} f(x) & \text{if \(f(x) \in U_p\)} \\ \displaystyle f(a) + t\left[tr + (1-t)f\left(a + t\left(\frac{b-a}{2} - a\right)\right) - f(a)\right] & \text{if $\displaystyle x \in \Bigg[a, a + t \left[\frac{b-a}{2} - a\right]\Bigg)$} \\ t r + (1-t) f(x) & \text{if $\displaystyle x \in \Bigg[a + t \left[\frac{b-a}{2} - a\right], b -t\left[b - \frac{b-a}{2}\right]\Bigg]$} \\ \displaystyle t r + (1-t) f\left(b - t\left(b- \frac{b-a}{2}\right)\right) + t\left[f(b) - t r - (1 - t) f\left(b - t\left(b- \frac{b-a}{2}\right)\right)\right] & \text{if $\displaystyle x \in \Bigg(b - t\left[b - \frac{b-a}{2}\right], b\Bigg]$} \end{cases} $$ Let $i \colon [0, 1] \times K \rightarrow \rthree$ be a function defined by $i(t, f(x)) := i_t(f(x))$. My intention is to define $i$ such that it keeps the part inside $U_p$ the same and shrinks the outside part. The end points of the shrunk part and the part inside $U_p$ are joined linearly.
I am not able to prove that $i$ is an isotopy.
Assuming it is, $U_p \cap K$ is straight line isotopic to the line joining $f(a)$ and $f(b)$. Let this isotopy be $s$. $(s \circ i) (1, K)$ is a triangle with the vertices $r$, $f(a)$ and $f(b)$. This triangle is isotopic to $\mathbf{S}^1$.
How do I go about proving that $i$ is indeed an isotopy? I don't have trouble understanding that each $i_t$ is bjiective and continuous. The pre-image of an open set in $\mathbf{R}^3$ is equal to the pre-image of the intersection of the image of all $i_t$s with the open set. Now, I would have to prove that this pre-image is of the form $(A, f(C))$, where $A$ and $C$ are an open sets in $[0,1]$ and $[0, 2 \pi)$ respectively, right? $C$ is open iff $f(C)$ is open.
By local flatness, I mean the following. A point $p \in K$ is said to be locally flat if there exists a neighbourhood $U_p \ni p$ such that the ordered pair $(U_p, U_p \cap K)$ is homeomorphic to $(B, d)$, where $B \subset \mathbf{R}^3$ is the unit ball around origin and $d \subset B$ is the set of points along a diameter of the ball. A knot $K$ is said to be locally flat if all points are locally flat.
By isotopy, I mean a homotopy $i$ such that all the individual $i_t$s are bijective. I distinguish between isotopy and ambient isotopy, although people usually mean ambient isotopy when they mean isotopy in a knot theory context. By ambient isotopy, I mean $a \colon [0, 1] \times \mathbf{R}^3 \rightarrow \mathbf{R}^3$. The domain in the ambient isotopy is not $\mathbf{S}^1$ but rather $\mathbf{R}^3$. My terminology follows Knots and Links by Cromwell.
I'll collect what i wrote in the comments
Let $K\subset S^3$ be a knot that is locally flat at some point $p\in K$. Then let $U_p$ be the neighbourhood of $p$ s.t. $(U_p,U_p\cap K)\rightarrow (\mathbb{R}^3,\mathbb{R})$ is a homeomorphism of pairs.
Then let $B = S^3\setminus U_p \cong D^3$, $A= B\cap K$ and $\{a,b\}=\partial B \cap K$. Then $A$ is a possibly wild arc with endpoints $\{a,b\}$ whilst its complement $I=K\setminus A$ is a tame arc. Thus we may pick another 3-ball $B'$ which contains $B$ in its interior, intersects $I$ in two points $\{a',b'\}$ and so $ K\cap (B'\setminus B)= J_a\cup J_b$ where $J_a$ and $J_b$ are tame arcs with end points $a,a'$ and $b,b'$ respectively.
Consider homeomorphisms $h_i:B' \rightarrow B'$ that is constant on $\partial B'$ but shrinks everything in the interior by a factor of $\frac{1}{i}$ for $i\in (0,1]$ to some point i will call $0\in Int(B)\cap K$. Extend this by identity to $H_i:S^3\rightarrow S^3$.
Fix $K$ as subset of $S^3$ and let us look at what this map does to $K$. For $i=1$ it doesn't move $K$ at all but as $i\rightarrow 1$ the arcs $J_a$ and $J_b$ get stretched towards $0\in B$ so that in the limit we get that image of is a tame arc plus two straight line segments which are the images of $J_a$ and $J_b$ (this makes a tear shape as you say). Note that $a'$ and $b'$ stay fixed but $a$ and $b$ get closer and closer to $0$.
Let $f: S^1\rightarrow S^3$ be the embedding of the knot $K$. Then define an isotopy $I_i=H_i\circ f$ for $i\in (0,1]$ and $I_0(x)= \varinjlim I_i(x)$ for $x\in S^1\setminus f^{-1}(0)$ and $I_0(f^{-1}(0))= 0$ then $I_1$ is continuous and injective. Clearly all other $I_i$'s are also continuous and injective (and therefore embeddings).
Can you see why here continuity in the product holds? (An open set interestcs a simple closed curve in a countable union of open arcs - as the simple closed curve moves continuously the sizes of the arcs also change continuously).