Braid groups on topological spaces

Question

The configuration space $C_n(M)$ of $n$ particles in some connected graph $M$ (thought of as the topological realisation of a one-dimensional CW-complex) is $$M^n \backslash \{ (x_1, \ldots, x_n) \mid x_i=x_j \ \text{for some} \ i \neq j\},$$ and the corresponding unordered configuration space $UC_n(M)$ is the quotient $C_n(M)/ \mathfrak{S}_n$ where $\mathfrak{S}_n$ acts freely on $C_n(M)$ by permuting the coordinates. The braid group $B_n(M)$ is the fundamental group of $UC_n(M)$ (which is connected if $M$ is connected itself).

Is it true that $B_{m}(M)$ admits an injective homomorphism into $B_n(M)$ if $m \leq n$? Is it true at least if $m \leq 2$?

It is worth noticing that the question is not trivial if we replace the graph $M$ with a sphere according to this question.

Answer 1

I think the answer is yes if $M$ has a leaf. Subdivide $M$ sufficiently so that Abram's locally CAT(0) model $X_n(M)$ of $B_n(M)$ (called the reduced braid group, $RB_n(M)$ is this paper of Crisp and Wiest) can be used. We can map $X_n(M)$ to $X_{n+1}(M')$ by mapping a configuration $(p_1, \dots, p_n)$ to $(p_1, \dots, p_n, p_{n+1})$. Here $M$ is the subdivided graph and $M'$ is $M$ with an additional subdivided edge glued onto the degree one vertex of the distinguished leaf of $M$. The point $p_{n+1}$ is fixed and corresponds to the degree one vertex of the distinguished (extended) leaf of $M'$. The map $X_n(M) \to X_{n+1}(M')$ is a cubical map and it seems that the image of the link of any vertex in $X_n(M)$ is a full subcomplex of the link of the image point in $X_{n+1}(M')$. So, this map is $\pi_1$-injective. But, $X_{n+1}(M')$ is isomorphic to $X_{n+1}(M)$, so we have an injective homomorphism $B_n(M) \to B_{n+1}(M)$.

Here's some additional detail/background, though Crisp & Wiest (and Abrams in his thesis) say it much better if you read their work. Given a subdivided graph $M$, the carrier of a point $x$ in $M$ is the closed cell (vertex or edge) containing it. The space $Y_n(M)$ consists of configurations of $n$ ordered points in $M$ such that the carriers of points are mutually disjoint. $X_n(M)$ is the quotient under the action of the symmetric group. The cubical structure comes from the fact that a configuration of $n$ points where exactly $k$ of these points have edges as their carriers corresponds to a $k$-cube.

Answer 2

When $M$ is the 2-sphere, then there is no an injective group homomorphism from $B_{n}(S^2)$ to $B_{m}(S^2)$ if $m$ is greater than $n$. Indeed, $\sigma_1\sigma_2\dots \sigma_{n-1}^2\dots\sigma_2\sigma_1=1$ is trivial in $B_{n}(S^2)$ but not a relation in $B_{m}(S^2)$ so there is no group homomorphism from $B_{n}(S^2)$ to $B_{m}(S^2)$.