# All finite path connected CW complexes are homotopy equivalent to a CW complex with only one 0-cell.

Statement: if $$X$$ is a finite path connected CW-complex, then $$X$$ is homotopy equivalent to a CW-complex with only one $$0$$-cell.

Thoughts: we can use the theorem that for a contractable subspace $$A \subseteq X$$ such that $$(X, A)$$ has the homotopy extension property, the projection $$X \to X/A$$ is a homotopy equivalence. I want to apply this for each $$1$$-cell that does not have its ends glued together. This is ok because then it is a contractable, there are only finitely many, so eventually this procedure will stop and $$(X, A)$$ is a relative CW-complex, so it has HEP.

We will be left with only $$1$$-cells that have their ends glued together. I'd like to conclude from path connectedness that there is only one $$0$$-cell left, but could not manage this.

Consider the $$1$$-skeleton $$X^1$$ of $$X$$. Since $$X$$ is path connected, the cellular approximation theorem shows that $$X^1$$ is path connected. We have finitely many $$0$$-cells $$\{x_i\}$$, $$i = 1,\ldots,m$$. Let us show how to reduce the number of $$0$$-cells by $$1$$ (if $$m > 1$$). There exists a path from $$x_m$$ to $$x_1$$. Hence there must exist a closed $$1$$-cell $$e^1$$ attached to $$\{x_m\}$$ and some $$\{x_i\}$$ for $$i < m$$. Clearly $$e^1$$ is a contractible subcomplex of $$X$$ (with $$0$$-skeleton $$\{x_m,x_i\}$$) , thus the quotient map $$X \to X' = X/e^1$$ is a homotopy equivalence. The space $$X'$$ is a path connected CW-complex. The number of $$0$$-cells is $$m-1$$. Thus, proceeding inductively, we get the desired CW-complex with one $$0$$-cell.
In fact $$X^1$$ is a multigraph, i.e. a graph which is permitted to have multiple edges between two vertices (note that edges can be self-loops connecting a vertex to itself). It is well-known that each connected multigraph has a spanning tree which contains all vertices. Recall that a tree is a contractible subgraph. See for example Hatcher's "Algebraic Topology" Proposition 1A.1. So let $$T$$ be a spanning tree of $$X^1$$. It is a subcomplex of $$X$$, thus the quotient map $$X \to X/T$$ is a homotopy equivalence. The CW-complex $$X/T$$ has only one $$0$$-cell.
If $$X$$ is a CW-complex, then the inclusion $$i:X^1\to X$$ of the $$1$$-skeleton induces a bijection $$\pi_0(X^1)\to \pi_0(X)$$. Indeed, $$X$$ is obtained from $$X^1$$ by attaching cells of degree $$2$$ and higher, which do not affect $$\pi_0$$ (this follows from cellular approximation, but you can see it really concretely: if $$n\geq 2$$ then $$S^{n-1}$$ is connected so each new cell just gets attached to one of the connected components you already had).
In particular, if $$X$$ is connected, then its $$1$$-skeleton $$X^1$$ is connected. If all edges in $$X^1$$ start and end at the same vertex, then it is clear that each vertex together with all its edges forms a connected component of $$X^1$$, so if $$X^1$$ is connected it can only have one vertex.