I'm looking at various seminars that Pascal Lambrechts has given about manifold calculus. One of the lectures I've found is in the following slides of one of his lectures, but I have problems understanding it.
In particular, I'm struggling to understand how one can achieve the result in this slide by the proof shown in it:
Here, $O_1$ and $O_2$ represent half of the northern and southern hemispheres and a bit more on the sides. They are chosen in a way that $O_1\cap O_2$ can be identified as the disjoint union of two intervals (or $\Bbb D^1\cong [-1,1]$). Below, I use the notation of $A$ and $B$, which correspond respectively to $O_1$ and $O_2$. Here's a graph that shows each element:
I just need to understand it for $V = \Bbb R^d$, so we would have the theorem
$\operatorname{Imm}(\Bbb S^1,\Bbb R^d) \simeq \operatorname{Map}(\Bbb S^1,\Bbb S^{d-1})$
with the following diagram:
I've tried to understand the result for $d=2$, in which we can compute the pullback of $\Bbb S^1\stackrel{\Delta}{\longrightarrow} \Bbb S^1\times\Bbb S^1 \stackrel{\Delta}{\longleftarrow} \Bbb S^1$. I don't know what $\Delta$ is, but I considered the diagonal of $\Bbb S^1$: $$\Delta:\Bbb S^1\to\Bbb S^1\times\Bbb S^1,\quad x\mapsto(x,x).$$
In this case, the pullback would be $$\Bbb S^1\times_{\Delta\times\Delta}\Bbb S^1 = \{(x,y)\in\Bbb S^1\times\Bbb S^1 : (x,x) = \Delta(x) = \Delta(y) = (y,y)\} = \{(x,x)\in\Bbb S^1\times\Bbb S^1\}.$$
But in this case, the pullback is just the diagonal of the torus, which is just a copy of $\Bbb S^1$. I don't know how to find a homotopy equivalence (or something mor powerful) from $\Bbb S^1$ to $\operatorname{Map}(\Bbb S^1,\Bbb S^1)$ to get what I'm looking for.
Could anyone please help me out understanding the proof of the Smale's theorem? Or at least, understand the case for $d = 1$. Thanks in advance!
Replace $\Delta:X\to X\times X$ by the fibration $X^I\to X\times X$ given by evaluation at the two endpoints. Then, a point in the pullback of the new square is given by a pair of paths $p,q: I\to X$ so that $p(0)=q(0)$ and $p(1)=q(1)$, these paths can be glued together at the ends to make free loops $S^1\to X$. This is the idea, but you may need to dig into homotopy limits/model categories a bit to understand this.