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Let $X_1$ and $X_2$ be two cones over the hawaiian earring and let $X$ be the wedge sum of $X_1$ and $X_2$ (of course you join them in the special point of the hawaiian earring). How do you prove that $X$ is not contractible? Thanks

flag	asked Dec 21 at 15:46 Pedro Perez 49●1

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@Pedro: If this is a homework, then MO is not an appropriate place for the question, try math.stackexchange instead. If this is not a homework, you should give a motivation for the question, explain why do you think the space is not contractible, what did you try to solve this problem, etc. – Misha Dec 21 at 18:06

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Also: Try to search "wedge sum" on MO; the first hit you see will get you started on the problem regardless of its origin. – Misha Dec 21 at 18:18

Henry T. Horton · Answer 1 · 2012-12-21 19:51:44Z UTC

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See the proof of Theorem 2.6 of

J.W. Cannon, G.R. Conner, The combinatorial structure of the Hawaiian Earring group, Topology Appl. 106 (3) (2000) 225–271.

A copy of this paper is available on the second author's webpage here.

answered Dec 21 at 19:51

Henry T. Horton
301●1●3●4

Jeremy Brazas · Answer 2 · 2012-12-22 00:56:48Z UTC

The space you are talking about is sometimes called the Griffiths space. As Henry Horton suggests, to prove it is not contractible you can show the fundamental group is non-trivial (even though its Cech homotopy groups are trivial). The reference already given is a good one though I'll add that Griffith was the first to show this in

H. B. Griffiths, The fundamental group of two spaces with a common point, Quart. J. Math. 5 (1954), 175-190.

Another discussion on non-contractible one-point unions of simply connected spaces is in

K. Eda, A locally simply connected space and fundamental groups of one point unions of cones, Proc. Amer. Math. Soc. 116 No. 1 (1992) 239-249.

Is the wedge sum of two cones over the hawaiian earring contractible?

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