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I've been asked which of A,B,C...,Z are contractible.

Intuitively, I can see that all but A, B, D, O, P, Q and R are contractible. These are all (except B) homotopy equivalent to a circle, and B homotopy equivalent to 'two circles' (what do I actually mean by this?).

Is this formal enough? How could I make it more formal, if not? How do I show that a circle isn't contractible?

Thanks

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It's a little tricky to answer this question without knowing what you're allowed to assume (that is, what's already been proved in the class and what hasn't). If I were asking this question on a homework, I probably wouldn't be expecting a formal answer, instead the point is just to test that you intuitively understand the concept. –  Noah Snyder Oct 13 '11 at 16:09
    
In answer to your question about the letter B, it is homotopy equivalent to the wedge of two circles, which we denote by $S^1\vee S^1$. It is not contractible either because $\pi_1(S^1\vee S^1)\cong\mathbb Z*\mathbb Z$, where the $*$ denotes the free product. –  SL2 Oct 13 '11 at 16:11

1 Answer 1

By definition, a contractible space must have trivial homotopy groups, it must have trivial fundamental group, i.e. be simply connected. However, a standard result shows that fundamental group of the circle is not trivial, in fact, it's isomorphic to $\mathbb{Z}$. From this you get your intuition that letters with a circle are not contractible, and we can use van Kampen's theorem to compute the fundamental group of the wedge sums.

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