Let $\pi\_{1}: \text{Top}^\* \rightarrow \text{Grp}$ denote the fundamental group functor and let $H\_{1}: \text{Top}^\* \rightarrow \text{Grp}$ denote the first homology group functor. We can then define a natural transformation $\eta: \pi\_{1} \rightarrow H\_{1}$ with a component $\eta\_{X}: \pi\_{1}(X) \rightarrow H\_{1}(X)$. We define the configuration space on $n$ points by,
$$C\_{\hat{n}}(\mathbb{R}) = \{(z\_{1},...,z\_{n}) \in \mathbb{R}^n \; | \; z\_{i} \neq z\_{j}\}$$
and the unordered configuration space on $n$ points by $C\_{n}(\mathbb{R}) = C\_{\hat{n}}(\mathbb{R}) \big/ \Sigma\_{n}$, where $\Sigma\_{n}$ is the symmetric group. We then have for $\vec{p} \in C\_{\hat{n}}(\mathbb{R})$ that $\pi\_{1}(C\_{\hat{n}}(\mathbb{R}),\vec{p}) \cong PB\_{n}$ and $\pi\_{1}(C\_{n}(\mathbb{R}),\tau(\vec{p})) \cong B\_{n}$, where $\tau : C\_{\hat{n}}(\mathbb{R}) \rightarrow C\_{n}(\mathbb{R})$ is the orbit space projection, $PB\_{n}$ is the pure braid group on $n$-strands, and $B\_{n}$ is the braid group on $n$-strands.


Consider the components of the natural transformation between the fundamental group functor and the first homology group functor given by,


$$\eta\_{C\_{\hat{n}}(\mathbb{R})} : \pi\_{1}(C\_{\hat{n}}(\mathbb{R})) \rightarrow H\_{1}(C\_{\hat{n}}(\mathbb{R}))$$
$$\eta\_{C\_{n}(\mathbb{R})} : \pi\_{1}(C\_{n}(\mathbb{R})) \rightarrow H\_{1}(C\_{n}(\mathbb{R}))$$



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> What objects in $\text{Grp}$ will these components of $\eta: \pi\_{1} \rightarrow H\_{1}$ associate with $PB\_{n}$ and $B\_{n}$, and how do they relate to $C\_{\hat{n}}(\mathbb{R})$ and $C\_{n}(\mathbb{R})$?
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I am interested in this question because I am wondering if $H\_{1}(C\_{\hat{n}}(\mathbb{R}))$ and $H\_{1}(C\_{n}(\mathbb{R}))$ have any unnoticed connections to knot theory since every knot is the closure of a braid.



The First Homology Group of Configuration Space and Knot Theory
In line with Richard Kents answer, what you obtain by passing from the fundamental group to the first homology group of the ordered configuration spaces should be generated by letting any two strands in the pure braid group tangle together - and allow for no further tangling of three and higher strands. You should be able to make this clear by considering the morphism $\gamma\_n \colon C\_{\hat{n}}(\mathbb{R}^2) \to \prod\_{n \choose 2} C\_{\hat{2}}(\mathbb{R}^2)$ picking any two configurations of points in the domain and mapping them onto the configuration indexed by this choice of two points.


The morphisms $\gamma\_{n-1}$ and $\gamma\_n$ fits as a morphism of vector bundles between the bundle


$$
\bigvee\_{n-1} S^1 \to C\_{\hat{n}}(\mathbb{R^2}) \to C\_{\hat{n-1}}(\mathbb{R^2})\\ 
$$


and the trivial bundle


$$
\prod\_{n-1} C\_{\hat{2}}(\mathbb{R}^2) \to \prod\_{{n} \choose 2} C\_{\hat{2}}(\mathbb{R}^2) \to \prod\_{{n-1} \choose 2} C\_{\hat{2}}(\mathbb{R}^2)
$$


(Pascals triangle tells you that the indices in the fibers indeed match up) - and using the homotopy equivalence $C\_{\hat{2}}(\mathbb{R}^2) \simeq S^1$, you can see the fibers of the upper bundle as the 1-skeleton of the $(n-1)$-tori in the fiber of the lower bundle, inductively meaning that $\gamma\_n$ becomes an $H\_1$-isomorphism for all $n$.


