# 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!

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Simiplicial approximation should give it to you immediately. Loops are $1$-dimensional, so can be pushed into the $1$-skeleton and homotopies are $2$-dimensional, so can be pushed into the $2$-skeleton. You can also use Seifert-VanKampen's theorem. –  Grumpy Parsnip Jun 2 '12 at 18:06
@JimConant: I agree that it should be immediate, but I can't see how to formalise your suggestion of pushing into. Could you possibly expand on this? Many thanks! –  Edward Hughes Jun 2 '12 at 18:31
"Pushing into" means that there is a homotopy. So, every loop is homotopic to a loop in the $1$-skeleton. Then you need to argue that if two loops are homotopic then they are homotopic through the $2$-skeleton. The SAT implies that a homotopy regarded as a map from $I^2$ is itself homotopic to one that lies in the $2$-skeleton. (However the fact that the homotopy is homotopic to the new one is not relevant for the argument.) –  Grumpy Parsnip Jun 2 '12 at 18:49
Okay - I'll try to work through that idea! –  Edward Hughes Jun 2 '12 at 19:12
yes, that's what I was thinking. –  Grumpy Parsnip Jun 3 '12 at 23:52