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Let M be a connected topological manifold of dimension $\geq2$ and let $M^n=M\times\dots\times M$ be the product on $n\geq1$ copies of M with the product topology. Set $\mathcal{F}_n(M)=\{(u_1,u_2,\dots,u_n)\in M^n\mid u_i\neq u_j, \forall i\neq j\}$. Consider the quotient space $\mathcal{C}_n(M)=\mathcal{F}_n(M)/\mathfrak{S}_n$, where $\mathfrak{S}_n$ is the symmetric group on $n$ letters. The group $\pi_1(\mathcal{C}_n(M))$ is called the braid group of M.

I know that for $M=\mathbb{R}^2$, $\mathcal{B}_n\cong\pi_1(\mathcal{C}_n(\mathbb{R}^2),q)$ where $q$ is the point represented by the unordered set $\{(1,0),(2,0),\dots,(n,0)\}$, and $\mathcal{B}_n$ is Artin's braid group.

I want to solve the following problem which states:

"Let $U\subset \mathbb{R}^2$ be an open disk. Prove that the inclusion homomorphism $\pi_1(\mathcal{C}_n(U),q)\longrightarrow\pi_1(\mathcal{C}_n(\mathbb{R}^2), q)$ is an isomorphism for any $q\in\mathcal{C}_n(U)$."

I've tried but I am unable to solve it. Could anyone help me please? I would be very grateful to you.

William
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    Try to show that the inclusion map $(C_n(U) ,q) \to (C_n(\mathbb{R}^2),q)$ is a pointed homotopy equivalence – William Mar 26 '20 at 17:53
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    First show that there is an ambient isotopy $H\colon\mathbb{R}^2 \times I \to \mathbb{R}^2$ such that $H_1$ is a homeomorphism $\mathbb{R}^2 \cong U$, and such that $H$ fixes the elements of $q$ for all time (first do it for the case where $U$ is centred at the origin because it's easier). If $i\colon U \to \mathbb{R}^2$ is the inclusion then $i\circ H_1 \sim id_{\mathbb{R}^2}$ and $H_1 \circ i \sim id_U$ through isotopies which fix $q$. Try to use $H$ to construct a homeomorphism $C_n(\mathbb{R}^2) \to C_n(U)$ which is a homotopy-inverse of $C_n(U) \to C_n(\mathbb{R}^2)$. – William Mar 26 '20 at 19:28
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    You have to be careful to make sure things are injective where necessary, because if $f\colon X \to Y$ is not injective then it won't induce a map on configuration spaces. – William Mar 26 '20 at 19:30
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    I think "ambient isotopy" is also not quite what I meant. Specifically I mean you should construct $H$ so that $H_0 = id_{\mathbb{R}^2}$ and for each $t$ $H_t$ should be a topological embedding. (for an ambient isotopy, every $H_t$ is required to be a homeomorphism $\mathbb{R}^2 \cong \mathbb{R}^2$, which is not what you should do) – William Mar 26 '20 at 22:59

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I will give an outline of the argument I had in mind from the comments, but leave some details for you to fill in yourself. I ended up taking a slightly different approach (blowing $U$ up instead of shrinking $\mathbb{R}^2$ down) because it felt more direct.

I'll just consider the case where $U=B_\frac{\pi}{2}(0)$, the open ball of radius $\frac{\pi}{2}$ centred at the origin (you should figure out how to adapt this argument/result to the general case). If we use polar coordinates then we can define a homeomorphism $\tau\colon U \to \mathbb{R}^2$ by $\tau (r, \theta) = (tan(r), \theta)$.

However, we want everything to preserve basepoints so we need to fix the elements of $q$. For this purpose choose a closed disk $D\subset U$ centred at the origin with radius $\rho< \frac{\pi}{2}$ such that all of the elements of $q$ are contained in $D$. Now choose a bump function $\psi \colon \mathbb{R} \to I$ such that $\psi((-\rho, \rho)) = 1$ and $\psi((-\frac{\pi}{2},\frac{\pi}{2})^c) = 0$, and define a new homeomorphism $h \colon U \to \mathbb{R}^2$ by $$ h(r,\theta) = \big(r + (1 - \psi(r))\cdot tan(r),\theta\big)$$ (you can see that this function is injective because it is strictly increasing in the radial coordinate). Then $h$ fixes $D$ pointwise, and there is an isotopy $\iota_U \sim h$ (where $\iota_U$ is the inclusion) given by

$$ H(r,\theta;s) = \big(r + s(1 - \psi(r))\cdot tan(r),\theta\big)$$ which fixes $D$ for all time.

Now let's turn to configuration spaces. If $f\colon X \to Y$ is an injective continuous function then it induces a continuous function $C_n(f) \colon C_n(X) \to C_n(Y)$, by applying $f$ coordinate-wise and noting it is well-defined because you quotient out by the same action on both sides. (Note that if $f$ were not injective then applying it coordinate-wise would not send $F_n(X)$ to $F_n(Y)$.) Moreover if $f$ is a homeomorphism then so is $C_n(f)$.

In particular, for each $s$ our $H_s$ above induces a pointed map $(C_n(U), q)\to (C_n(\mathbb{R}^2), q)$ and these assemble into a homotopy $\tilde{H}\colon C_n(U) \times I \to C_n(\mathbb{R}^2)$ where $\tilde{H}_0 = C_n(\iota_U)$ is the inclusion we're interested in, $\tilde{H}_1 = C_n(h)$ is a homeomorphism, and $\tilde{H}$ fixes $q$ for all time. I.e. our inclusion of configuration spaces is pointed-homotopic to a homeomorphism (in fact it is a pointed homotopy equivalence, where the homotopy inverse is $C_n(h^{-1})$) and therefore the map on braid groups $(\iota_U)_*\colon \pi_1(C_n(U), q) \to \pi_1(C_n(\mathbb{R}^2),q)$ is an isomorphism.

(Caution: we cannot formally say that "$\tilde{H} = C_n(H)$" because it is not the case that $C_n(X\times I) \cong C_n(X) \times I$. You will have to argue more directly that $\tilde{H}$ is continuous with respect to $s$.)

Edit: The above argument seems to generalize to the following statement, which might be enlightening for you to prove:

Suppose $f\colon X \to Y$ is a topological embedding which is isotopic to a homeomorphism, through an isotopy that fixes a set $q\subset X$ of $n$ points. Then the induced function $C_n(f)\colon (C_n(X),q) \to (C_n(Y),q)$ is a pointed homotopy-equivalence.

William
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