Does the pair $(S^n, \{p_1,p_2, \cdots, p_k\})$ have HEP?

Question

Let $p_1, p_2, \cdots, p_k$ be $k $-distinct points on $S^n.$ Then does the pair $(S^n , \{p_1,p_2, \cdots , p_k\})$ have homotopy extension property?

I am actually trying to show that $(S^n , \{p_1,p_2, \cdots , p_k\})$ is a CW pair but I am unable to figure out the underlying CW complex structure. A small hint is very much required at this stage.

Thanks!

Answer 1

You can be very explicit in the case of points with how to get HEP, but if you just care that it exists you can appeal to the fact that CW pairs have HEP. Given any set of points $\{p_1,p_2,\dots,p_k\}$ in a CW complex $X$, we may always subdivide the cells that each point lie in to make each $p_i$ a 0-cell .

Answer 2

You asked for an explicit construction in the comments of Connor Malin's answer. Let me provide one. We will give a CW-structure of the $n$-sphere with $0$-skeleton given by $k$ points $x_1,\ldots,x_k$. Note that this will imply the $(S^n,\{p_1,\ldots,p_k\})$ is a CW-pair since we can just choose a homeomorphism $S^n\to S^n$ mapping each $x_i$ to $p_i$ yielding a CW-structure for $S^n$ with $0$-skeleton $\{p_1,\ldots,p_k\}$.

We start with the $0$-skeleton $x_1,\ldots,x_k$. We then attach $k$ $1$-cells along the pairs of points $(x_i,x_{i+1})$ and $(x_k,x_1)$. The $1$-skeleton then becomes $S^1$. We can view this as the equator of a $2$-sphere. Indeed, we attach two $2$-cells corresponding to the upper and lower hemisphere. The $2$-skeleton then becomes $S^2$. This can then be viewed as the equator of a $3$-sphere, and we can attach two $3$-cells to obtain $S^3$. This process can be iterated until you get $S^n$.