The Fundamental Group of a Polyhedron Depends Only on its $2$-Skeleton Does anyone have a good quick proof of this using the Simplicial Approximation Theorem? I'm aware that it comes out as a corollary when considering edge paths and the edge group, but this seems like quite heavy machinery for what should be a simple idea. I haven't been able to put together a convincing argument myself though! 
 A: Let $X$ be a polyhedron, and consider the inclusion $i : \operatorname{sk}_2 X \to X$. It induces a morphism on fundamental groups $i_* : \pi_1 \operatorname{sk}_2 X \to \pi_1 X$, for some choice of base point $x_0$ in the $0$-skeleton. This morphism is:


*

*Surjective: let $\alpha : I \to X$ be a representative of some class in $\pi_1 X$, $\alpha(0) = \alpha(1) = x_0$. Then by the simplicial approximation theorem, since $I$ is 1-dimensional, then $\gamma$ is homotopic to $\beta : I \to \operatorname{sk}_2 X$; since $\alpha$ was already simplicial on the 0-skeleton of $I$, we can assume the homotopy is constant on the 0-skeleton, hence $\beta(0) = \beta(1) = x_0$ too. Then $i_*[\beta] = [\alpha]$.

*Injective: let $\alpha, \beta : I \to \operatorname{sk}_2 X$ be two loops such that $i_*[\alpha] = i_*[\beta]$, i.e. there is a homotopy $H : I \times I \to X$ such that $H(t,0) = \alpha(t)$, $H(t,1) = \beta(t)$, and $H(0,s) = H(1,s) = x_0$. Then similarly by the simplicial approximation theorem, $H$ is homotopic to some $G : I \times I \to \operatorname{sk}_2 X$ ($I \times I$ is 2-dimensional), and one can moreover take the homotopy to be constant on $\partial(I \times I)$ (on which $H$ is already simplicial). Then $G$ is a homotopy in $\operatorname{sk}_2 X$ between $\alpha$ and $\beta$, hence $[\alpha] = [\beta]$.


It follows that $i_* : \pi_1(\operatorname{sk}_2 X) \to \pi_1(X)$ is an isomorphism. More generally one can immediately adapt the arguments above to show that $\pi_n(\operatorname{sk}_{n+1} X) \cong \pi_n(X)$.
