Is “factoring through a dendrite loop” preserved under deletion? Definition 1. A dendrite $X$ is a 1 dimensional retract of the closed unit disk. Equivalently, $X$ is a compact, locally path connected, and uniquely arcwise connected metric space.
Definition 2. The path $\alpha :[0,1]\rightarrow l_{2}$ “factors through a dendrite loop” if there exists a dendrite $X$, a loop $\gamma :[0,1]\rightarrow X$ (with $\gamma (0)=\gamma (1)$), and a map $\pi :X\rightarrow l_{2}$ such that $\alpha =\pi \gamma$.
Question. Suppose both the path $\alpha :[0,1]\rightarrow l_{2}$ and also the concatenated path $\alpha \ast \beta :[0,1]\rightarrow l_{2}$ can be factored through dendrite loops. Can $\beta $ be factored through a dendrite loop?

The following comments are intended to absorb some of the difficulties, and point to where the remaining difficulties lie.
The image of $\alpha$ can be an arbitrary Peano continuum embedded in Hilbert space $l_{2}$. However, all the difficult issues are present in the special case of the unit disk or unit cube.
Declare $\alpha :[0,1]\rightarrow l_{2}$ “trivial” if $\alpha $ factors through a dendrite loop. An elementary argument shows if the trivial loops $\alpha _{n} \rightarrow \alpha $ uniformly then $\alpha $ is trivial, i.e. the trivial loops in $l_{2}$ are closed in the uniform topology.
(Apply Ascoli's theorem and the fact that each trivial loop extends to a map of the unit disk, so that each component of each point preimage separates the disk into convex sets.)
Declare the path $\alpha $ to be “tree-trivial” if $\alpha $ lifts to a loop in a tree (a simply connected finite graph). An elementary proof shows $\beta $ is tree trivial if each of $\alpha $ and $\alpha \ast \beta $ is tree trivial.
An elementary proof shows if $\alpha $ is trivial then there exist tree-trivial loops $\alpha _{n}\rightarrow \alpha $.
(Lift $\alpha $ to a dendrite $X$ (a natural inverse limit of trees under strong deformation retraction, and project the retractions into $X$. We can even assume $X$ is an $R$-tree, i.e. a length space).)
Consequently the question at hand is equivalent to the following: Suppose $\alpha $ and $\alpha \ast \beta $ are trivial. Suppose $\alpha _{n}\rightarrow \alpha $ and $\alpha _{n}$ is tree trivial. Must there exist tree trivial loops $\alpha _{n} \ast \beta _{n}\rightarrow \alpha \ast \beta$?
Finally, it may be helpful consider the contrapositive, which can ultimately be reduced to the following:
Suppose the path $\alpha$ is irreducible, i.e. no subloop of the path $\alpha $ lifts to a dendrite loop. Suppose $\beta $ lifts to a dendrite loop. Can $\alpha \ast \beta $ lift to a dendrite loop?
 A: The answer to the original question is apparently no.
If a starting loop factors through a dendrite, and we delete a subloop which also factors through a dendrite, the surviving loop might not factor through a dendrite.
If we call a path irreducible if no nonconstant subloop factors through a dendrite, we can construct a counterexample to unique path lifting in the following context.
If $X$ is the $R$-tree of based irreducible paths in the plane, treating two paths as equivalent if they pass through the same points in the same order,
then endpoint projection $X \rightarrow R^{2}$ fails to have unique path lifting.
Partition the lower half of the closed unit disk into closed 0 cells, 1 cells, and 2 cells, so that each cell touches the simple closed curve boundary, the 1-cells touch at exactly their endpoints, and the 2 cells touch the boundary in 3 places, and so that no 0-cell touches the open lower semicircle.
Arrange also that the union of the 2-cells is dense in the horizontal, and that the 0-cells have full linear measure in the horizontal.
Topologically, the quotient space is a dendrite, and the boundary of the half disk determines a loop in the quotient space.
Now glue, (to the lower half disk) an upper closed half disk fibred by the canonical semicircles.
Generically, now we have obtained a partition of the closed unit disk into continua so that the resulting quotient space is 2 dimensional.
Delete the open semicircles from the upper open unit disk, but remember that $[-x,0]$ and $[0,x]$ are identified. Thus the horizontal bar of the closed unit disk determines a reverse pairing, the horizontal bar factors through an interval and particular a dendrite.
Deleting the mentioned reverse pair from the boundary of the closed upper semicircle leaves an injective loop which, in particular, cannot be lifted to a loop in a dendrite.
This unexpected answer was in fact hiding in plain sight. But it was easy to be tricked for 3 reasons:

*

*We get a yes answer for loops in 1 dimensional spaces.


*The property of ‘lifting to a dendrite loop’ is closed in the uniform topology for loops in any Peano continuum (with help from Ascoli's Theorem).


*Dendrites are similar to trees. Both are compact contractible locally contractible uniquely arcwise connected metric spaces.
But unlike a tree, endpoints of a dendrite are typically dense G-delta!
This fact is the real reason a counterexample can be built.
More details: Before gluing, the left and right sides of the lower unit disk determine dendrites. The 1 dimensional part of a dendrite (the complement of the endpoints) is F-sigma, a direct metric limit of trees. But the two F-sigma sets corresponding to the reverse pair (created after mating) will typically be disjoint. After deleting the newly created reverse pair, (in the new quotient space) we are left with a loop which is injective and which cannot be lifted to a loop in a dendrite.
Summary. We construct a cellular decomposition of the closed unit disk, so that antipodal points on $[-1,1]$ are equivalent. The restriction to the lower closed unit disk, and modding out by connected components, is a dendrite. On the horizontal bar $[-1,1]$, $x$ and $-x$ are equivalent, and thus restriction to the horizontal bar factors through a dendrite. No two points on the open lower semicircle are equivalent.
