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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.

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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.

Edited:

The finiteness assumption can be omitted, the assertion is true for any path connected CW-complex.

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.

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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.

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