Space of the trivial long knot in the thickened surface

Question

Let $F$ be a compact oriented surface and $x_0\in F$ a basepoint. Consider the set $\mathcal E=Emb_0(I,F\times I)$ of embeddings $\sigma\colon I\hookrightarrow F\times I$, $\sigma(\partial I)=\{x_0\}\times\partial I$, isotopic with fixed ends to the trivial long knot $\{x_0\}\times I$.

The natural projection $p\colon F\times I\to F$ maps a string $\sigma\in\mathcal E$ to a contractible loop $p\circ\sigma$ in $F$. On the other hand, the graphics $\Gamma_\omega\subset F\times I$ of a contractible loop $\omega\in\Omega(F,x_0)$ is isotopic to the trivial string. Thus, the space of contractible loops $\Omega_0 (F,x_0)$ is a retract of the space of long unknots $\mathcal E$.

My question is whether $\Omega_0 (F,x_0)$ is homotopy equivalent to $\mathcal E$.

In particular, this would imply that $\pi_i(\mathcal E)=\pi_i(\Omega_0 (F,x_0))=\pi_{i+1}(F)$, $i\ge 1$. For example, when $F=S^2$, $\pi_2(\mathcal E)=\pi_{3}(S^2)=\mathbb Z$, and the space of long unknots in the sphere is not trivial.

When $F$ is a disk, the answer is positive: by A. Hatcher the space of long unknots is contractible.

Answer 1

Let us show that $\mathcal E=Emb_0(I,F\times I)\sim\Omega_0(F,x_0)$.

We start with R. Budney's remark.

Proposition. Let $F$ be a connected compact 2-manifold and $P(F)$ the pseudoisotopy group, i.e the group of diffeomorphisms of $F\times I$ rel $\partial F\times I\cup F\times 0$. Then $P(F)$ is contractible.

Let $Diff(F\times I)$ be the diffeomorphism group of $F\times I$ rel $\partial(F\times I)$. From the exact sequence $$ 1\to Diff(F\times I)\to P(F)\to Diff_0(F)\to 1 $$ we have $Diff(F\times I)\sim \Omega Diff_0(F)$ and $Diff_0(F\times I)\sim \Omega_0 Diff_0(F)$. In particular, $Diff_0(S^2\times I)\sim \Omega_0(SO(3))$, and $Diff_0(F\times I)$ is contractible for other surfaces.

Consider the natural action of $Diff_0(F\times I)$ on the framed long knots. We have the exact sequence $$ 1\to Diff((F\setminus D)\times I)\to Diff_0(F\times I)\to \mathcal K^{fr}_0\to 1 $$ where $D\subset F$ is a small disk containing $x_0$ and $\mathcal K^{fr}_0$ is the orbit of the trivial (framed) unknot $x_0\times I$. Since $Diff((F\setminus D)\times I)\sim*$, we have $\mathcal K^{fr}_0\sim Diff_0(F\times I)\sim \Omega_0 Diff_0(F)$.

By forgetting the framing, we get the covering $p:\mathcal K_0^{fr}\to \mathcal E$.

1. Let $F=S^2$. There is a diffeomorphism $\psi\in Diff_0(S^1\times I)$ which maps the unknot $U=x_0\times I$ to itself and changes the framing. (Consider the isotopy $\psi_t$, $t\in[0,1]$ which pulls the unknot over the sphere so that the projection of $\psi(t,s)=\psi_t(x_0\times s)$ to $S^2$ realizes the generator of $\pi_1(\Omega_0(S^2,x_0))=\pi_2(S^2)=\mathbb Z$.) Then the fiber of $p$ is infinite (equals to $\mathbb Z$). Hence, for $i>1$ $$ \pi_i(\mathcal E)=\pi_i(\mathcal K_0^{fr})=\pi_{i+1}(SO(3))=\pi_{i+1}(S^3)=\pi_{i+1}(S^2)=\pi_i(\Omega_0(S^2,x_0)) $$ and $$ 1\to\pi_1(\mathcal K_0^{fr})\to\pi_1(\mathcal E)\to \mathbb Z \to 1. $$ Since $\pi_1(\mathcal K_0^{fr})=\pi_2(SO(3))=0$, $\pi_1(\mathcal E)=\mathbb Z=\pi_1(\Omega_0(S^2,x_0))$. The space $\Omega_0(S^2,x_0)$ is a retract of $\mathcal E$ and has the same homotopy groups, thus, $\mathcal E\sim\Omega_0(S^2,x_0)$.

2. Let $F\ne S^2$. Then $\mathcal K^{fr}_0\sim *$. We claim that the forgetting map $p:\mathcal K^{fr}_0\to \mathcal E$ is an equivalence. Let $U^{fr}_0$ be the unknot $x_0\times I$ with the initial framing, and $U^{fr}_1\in p^{-1}(U)$. Then there is an isotopy of long framed knots $U^{fr}_t$, $t\in[0,1]$. Consider the covering $\tilde F$ of $F$ which embeds in $\mathbb R^2$. Then the isotopy $U^{fr}_t$ lifts to $\tilde U^{fr}_t$ in $\tilde F\times I\subset\mathbb R^2\times I$. Hence, the framings of $\tilde U^{fr}_0$ and $\tilde U^{fr}_1$ (thus, of $U^{fr}_0$ and $U^{fr}_1$) coincide. Then $p^{-1}(U)=\{U^{fr}_0\}$ and $\mathcal E\simeq\mathcal K^{fr}_0\sim *\sim \Omega_0(F,x_0)$.