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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.

A counterexample is statistically guaranteed if one performs the following construction, generically, randomly, or in general position.

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:

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

2. 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).

3. 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.

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:

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

2. 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).

3. 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.

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.

A counterexample is statistically guaranteed if one performs the following construction, generically, randomly, or in general position.

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:

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

2. 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).

3. 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.

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.

A counterexample is statistically guaranteed if one performs the following construction, generically, randomly, or in general position.

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:

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

2. 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).

3. 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.

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.

A counterexample is statistically guaranteed if one performs the following construction, generically, randomly, or in general position.

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, and deleting the reverse pair from the 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:

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

2. 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).

3. 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.

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.

A counterexample is statistically guaranteed if one performs the following construction, generically, randomly, or in general position.

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:

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

2. 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).

3. 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.