Proposition: if $X$ is a simply-connected CW complex, it is homotopy equivalent to a complex $Y$ with trivial $1$-skeleton (i.e. $Y^1 = Y^0 = \{*\}$)
Proof sketch: Take a maximal tree $T$ in the 1-skeleton of $X$, and show $X\sim X/T$. So we can suppose the 0-skeleton is trivial. Compute the fundamental group of $X$ by noticing it is the same as the fundamental group of the 2-skeleton, by cellular approximation. Van Kampen then gives $$\pi_1(X^2) = \langle \text{ 1-cells } | \text{ boundaries of 2-cells }\rangle$$. Thus, every 1-cells is a word in the boundary of the 2-cells. In particular, attaching a cell with boundary the given 1-cell $e$ gives the complex $X/e$ (contract $e$ through the attached cell. On the other hand, being homotopic to a word in the boundary, i.e. nullhomotopic, shows that attaching this cell should be equivalent to wedging on a new 2-sphere, $$ X/e \sim X\vee \mathbb{S}^2 $$ Then killing the new cell gives $$X/e \cup \mathbb{D}^3 \sim X$$
Question: I would like to see this more directly. I want to say `look at the boundary of any 2-cell. This is nullhomotopic, so the complex is homotopic to one in which we just wedged this sphere on. Repeat until any interesting attachments are gone, then conclude there are no 1-cells by simple connectivity.' Unfortunately this fails, the attaching map is nullhomotopic in the 2-skeleton, but I need it to be nullhomotopic in the complex without the 2-cell added.
Can anyone give a direct argument? Also feel free to critique my sketch if there's something fishy…
Best Answer
Here's a slightly different idea (which might be almost the same idea; I'm a bit unsure about your construction).
From what you've written we can start by assuming that the $0$-skeleton is single vertex $v$.
Every $1$ cell $e$ is a closed loop with both endpoints at $v$. Since $X$ is simply connected, there exists a continuous map $h_e : D^2 \to X^{(2)}$ such that the restriction $h_e : S^1 \to e$ is a homeomorphism. Now attach a new 2-cell and 3-cell $D^2_e$, $D^3_e$, as follows. The attaching map of $D^2_e$ is a homeomorphisms onto $e$. Consider attaching map of $D^3_e$, which is a continuous map $S^2 \mapsto X^{(2)}$: the restriction of this attaching map to the upper hemisphere $H^+_e$ is a homeomorphism onto $D^2_e$, and the restriction to the lower hemisphere $H^-_e$ factors as a homeomorphism $H^-_e \mapsto D^2$ followed by the continuous map $D^2 \xrightarrow{h_e} X^{(2)}$.
Attaching these cells $D^2_e$ and $D^3_e$ for each $1$-cell $e$ of $X$, the resulting complex $\widehat X$ is pretty clearly homotopy equivalent to $X$. In fact there is a strong deformation retraction $f : \widehat X \to X$, such that the restriction $f \mid D^3_e$ is induced by a map on the $3$-disc which factors as a strong deformation restriction from the 3-disc to its boundary hemisphere $H^-_e$ composed with $h_e : D^2 \to X^{(2)}$.
Finally, take the quotient map $\widehat X \to Y$ where the union of the discs $D^2_e$ is collapsed to a single point; this quotient map is a homotopy equivalence. There's still some work to prove this, but it's sort of like the work you've already done to prove that collapsing a maximal tree in the original 1-skeleton is a homotopy equivalence.