Unordered configuration space of $\mathbb{R}P^1$

Question

In the paper

GEOMETRY OF TRUNCATED SYMMETRIC PRODUCTS AND REAL ROOTS OF REAL POLYNOMIALS, JACOB MOSTOVOY, Bull. London Math. Soc. (1998) 30 (2): 159-165,

Theorem 2. (b): $TP^n(\mathbb{R}P^1)$ is homeomorphic to $\mathbb{R}P^n$.

By page 161 line 2-line 3 of the above paper, the unordered configuration space of $k$ points on $\mathbb{R}P^1$, $C_k(\mathbb{R}P^1)=TP^k(\mathbb{R}P^1)\setminus TP^{k-2}(\mathbb{R}P^1)$.

My question: can we write $C_k(\mathbb{R}P^1)=TP^k(\mathbb{R}P^1)\setminus TP^{k-2}(\mathbb{R}P^1)=\mathbb{R}P^k\setminus \mathbb{R}P^{k-2}$?

(1). I note that $\mathbb{R}P^k\setminus \mathbb{R}P^{k-2}$ has only 3 cells of dimensions $k-2$, $k-1$, $k$.

On the other hand, by the paper

On the homology of configuration spaces, CF Bödigheimer, F Cohen, L Taylor, Topology, 1989,

(2). I obtain that $H^1(C_k(\mathbb{R}P^1);\mathbb{Z}/2)$ is non-trivial.

Question. (1), (2) are not compatible. Where is my mistake?

Answer 1

(1) and (2) are perfectly compatible, to the extent which (1) makes sense. $\mathbb{R}P^k$ is naturally a CW complex with one cell in each dimension, and $\mathbb{R}P^{k-2}$ is a subcomplex. Assuming that the inclusions are in the standard way, the difference $\mathbb{R}P^k \setminus \mathbb{R}P^{k-2}$ is not a union of open cells (of dimensions $k-1$ and $k$, but is not a CW complex. (Every CW complex must have at least one 0-cell, for instance.) Think about the case $S^{k-2} \subset S^k$ as an easier-to-visualize example: the difference $S^k \setminus S^{k-2}$ is homotopy equivalent to a circle, regardless of how large $k$ gets.