
Suspension homology
Let be a complex, and let , be two cones whose polytopes intersect only in .
The complex is called "SUSPENSION" of K.
Define: by the equation 
.
Show that $\phi$ induces a homomorphism:
.
By the way, the homology groups here are the reduced ones... Which has no importance for p>0.
[
What I've tried:
I thought I should define:
but what should I prove in order for it to be a homomorphism? it presereves the operation of chainaddition by definition of the cone operation...So what is left to do? can someone help me?
]
Thanks in advance

Re: Suspension homology
is obviously a homomorphism. To verify it induces a homomorphism between the homology groups, we need only to show ~ 0, that is maps boundary chains to boundary chains. This is only a straightforward computation:
It is easy to verify that , just write down and you can verify that.
So we have
~ 0, we're done

Re: Suspension homology
Hi xxx9,
Thanks for your answer. I've several questions regarding what you wrote:
1) What we've actually proved is that for all p>0. But what about the case of p=0? We don't need it because the reduced homology group of order 0 contains only the expression of ?
2) After we proved boundaries are copied into boundaries, shouldn't we prove it for cycles as well ? And then my will be obviously a homomorphism? If not, can you please explain me this last thing?
Hope you'll be able to help me with this
Thanks a lot again, you've been very helpful so far!

Re: Suspension homology
Actually what we have shown is for any , using this you can easily verify your 1) and 2), by just unwinding the definitions.

Re: Suspension homology
I still can't get it... The definition of the cone boundary operator for p=0 isn't the same as in the case of p>0 ... So we've actually shown it for p>0 .
From the fact that we can deduce that . Can you please explain why this is what we needed in order to prove ?
I really need your help in this
Thanks in advance

Re: Suspension homology
How do you define ?

Re: Suspension homology
?

Re: Suspension homology
NVM, I got it !
THanks a lot!