An **arc field** on a topological space $X$ is a continuous function $\Psi: X \rightarrow X^{[0,1]}$ such that for every $x \in X$, the path $\Psi(x): [0,1] \rightarrow X$   
(1) starts at $x$,   
 (2) is either an embedding (in other words a simple path), or the constant path on $x$. In the latter case we call $x$ a **singularity** of $\Psi$. Here $X^{[0,1]}$ is equipped with the compact-open topology.


This kind of definition that generalizes the notion of a vector field (or/and the tangent bundle) from smooth manifolds to the topological setting, goes back at least to Nash ([MR0071081](https://mathscinet.ams.org/mathscinet-getitem?mr=71081), A path space and the Stiefel-Whitney classes, 1955). There are results by Brown, Fadell, Stern ([MR0173269](https://mathscinet.ams.org/mathscinet-getitem?mr=173269), [MR0328952](https://mathscinet.ams.org/mathscinet-getitem?mr=328952), [MR0310880](https://mathscinet.ams.org/mathscinet-getitem?mr=310880)) which show that certain versions/corollaries of Hopf's index theorem generalize using arc fields. For example the following is true:


Suppose $X$ is either a compact topological manifold of dimension $\neq 4$, or a finite simplicial complex such that every punctured neighborhood $U - \{x\}$ is connected provided that $U$ is connected. If $X$ has zero Euler characteristic, it admits an arc field with no singularities.


My question is about moving the singularities around continuously when they are unavoidable. Feel free to assume $X$ is a compact smooth(able) manifold with nonzero Euler characteristic, but the more general the better of course. Write $\text{PConf}\_{k}(X) := \{(x\_{1}, \dots, x\_{k}): x\_{i} \neq x\_{j}\}$ for the $k$-fold ordered configuration space of $X$. Does there exist a finite $k$ and a continuous map $$\Theta: \text{PConf}\_{k}(X) \times X \rightarrow X^{[0,1]}$$ such that for every configuration $(x\_{1}, \dots ,x\_{k})$, the map $\Theta(x\_{1}, \dots, x\_{k}, - ): X \rightarrow X^{[0,1]}$ is an arc field whose singularities are contained in $\{x\_{1}, \dots, x\_{k}\}$?


I don't know the answer even for the sphere $S^2$, but have a convoluted way of showing that $k = 1$ won't do it. It feels like $k = 2$ should be enough for $S^2$, but I don't really know.



Continuously varying the singularities of a vector field
My answer will only address the smooth case, and vector fields rather than arc fields. Partly because I think your question is already interesting in this setting, but mainly because I'm not familiar with the topological case!


So assume $X$ is a closed smooth $n$-manifold with nonzero Euler characteristic. You ask whether there exists a finite $k$ and a continuous (or smooth) map
$$
\Theta: \operatorname{PConf}\_k(X)\times X\to TX
$$
such that $\Theta(x\_1,\ldots , x\_k,-):X\to TX$ is a vector field for each $(x\_1,\ldots , x\_k)\in \operatorname{PConf}\_k(X)$ whose zeroes are contained in $\{x\_1,\ldots ,x\_k\}$. Ignoring the condition on the zeroes, note that such a map $\Theta$ can be regarded as a section of the pullback bundle $\pi^\*TX$, where $\pi:\operatorname{PConf}\_k(X)\times X\to X$ is projection to last factor.


Now if the condition on the zeroes is satisfied, we get a nowhere zero section of the bundle $\pi\_{k+1}^\*TX$, where $\pi\_{k+1}:\operatorname{PConf}\_{k+1}(X)\to X$ is projection onto the final factor, aka the *Fadell-Neuwirth fibration*, see:


*Fadell, E.; Neuwirth, L.*, [**Configuration spaces**](http://dx.doi.org/10.7146/math.scand.a-10517), Math. Scand. 10, 111-118 (1962). [ZBL0136.44104](https://zbmath.org/?q=an:0136.44104).


This bundle has Euler class
$$
e(\pi\_{k+1}^\*TX)=\pi\_{k+1}^\*e(TX) \in H^n(\operatorname{PConf}\_{k+1}(X);\tilde{\mathbb{Z}}),
$$
where the tilde means coefficients may be twisted if $X$ is non-orientable. Here we have used naturality of the Euler class. If the Fadell-Neuwirth fibration $\pi\_{k+1}$ admits a section, then $\pi^\*\_{k+1}$ is injective on cohomology. Since $TX$ has non-vanishing Euler class, so does $\pi\_{k+1}^\*TX$, hence this bundle has no nowhere vanishing sections.


This happens for orientable surfaces of genus $\ge2$, for example. So I think in the smooth case, the analogous question about vector fields has a negative answer. I wouldn't be surprised if the answer is also negative for arc fields, for similar reasons.


I agree with your intuition about the $2$-sphere. I think you could also use the Euler class to rule out $k=1$, and it should be possible to construct $\Theta$ for $k=2$ just by deforming the usual vector field with two zeroes. 


