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## co-boundary in singular cohomology [closed]

Hi everyone. I'm a little stuck in the proof of d^2=0, being d the co-boundary operator of a singular n-simplex. I'd appreciate any help.

Thanks.

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A simplex has a boundary, not a coboundary... In any case, do the computation explicitly for $n=1$, $n=2$, $n=3$, and so on... Before $n$ gets too big, you will have seen the pattern. – Mariano Suárez-Alvarez Dec 22 2009 at 1:05
This really is a routine computation. Have you written out the first few cases? – David Speyer Dec 22 2009 at 1:06
This is a slight tangent, but does anyone know of a planar visual proof, or at least a proof by "planar diagrams" in the informal sense? When first trying to internalise this kind of calculation, I found myself stuck with what felt like a fiddly bit of hacking, or the lecturer's assurances of some geometric intuition that I didn't share. – Yemon Choi Dec 22 2009 at 2:19
Have you looked at Hatcher's "Algebraic Topology"? (see math.cornell.edu/~hatcher/AT/…). Assuming you do mean boundary and not coboundary, then it's Lemma 2.1 on page 105 of Hatcher's book (it's stated there not for singular homology, but the proof applies equally well to singular homology - singular homology is done in page 108). Is there a specific detail in the quick proof of Lemma 2.1 you are not comfortable with? – Vinoth Dec 22 2009 at 2:33
"I'm stuck, please help" with no reference is not really enough if the question is routine. People can only guess as to how to help without knowing what book you are reading and otherwise why you are stuck. – Greg Kuperberg Dec 22 2009 at 4:14
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