# Suspension homology

• Dec 16th 2011, 10:42 AM
WannaBe
Suspension homology
Let $\displaystyle K$ be a complex, and let $\displaystyle w_0 *K$ , $\displaystyle w_1 * K$ be two cones whose polytopes intersect only in $\displaystyle |K|$.
The complex $\displaystyle S(K)=(w_0 * K ) \cup (w_1 * K )$ is called "SUSPENSION" of K.
Define: $\displaystyle \phi : C_p(K) \to C_{p+1} (S(K))$ by the equation -
$\displaystyle \phi (c_p) = [w_o , c_p] - [w_1 , c_p ]$ .
Show that $\phi$ induces a homomorphism:
$\displaystyle \phi _{*} : H_p(K) \to H_{p+1} (S(K))$.

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:
$\displaystyle \phi _{*} (c_p + B_p(K) ) = \phi (c_p) +B_{p+1} (S(K))$
but what should I prove in order for it to be a homomorphism? it presereves the operation of chain-addition by definition of the cone operation...So what is left to do? can someone help me?
]
• Dec 16th 2011, 12:25 PM
xxp9
Re: Suspension homology
$\displaystyle \phi$ is obviously a homomorphism. To verify it induces a homomorphism between the homology groups, we need only to show $\displaystyle \phi(\partial{c})$ ~ 0, that is $\displaystyle \phi$ maps boundary chains to boundary chains. This is only a straightforward computation:
It is easy to verify that $\displaystyle \partial[w_0, c]=c-[w_0,\partial{c}]$, just write down $\displaystyle c=[v_0,...,v_{p+1}]$ and $\displaystyle \partial{c}=\sum_{k=0}^{p+1}(-1)^k [v_0,...,v_{k-1},v_{k+1},...,v_{p+1}]$ you can verify that.
So we have $\displaystyle \phi(\partial{c})=[w_0,\partial{c}]-[w_1,\partial{c}]$
$\displaystyle =(c-\partial[w_0,c])-(c-\partial[w_1,c])$
$\displaystyle =\partial([w_1,c]-[w_0,c])$
$\displaystyle =\partial(-\phi(c))$ ~ 0, we're done
• Dec 17th 2011, 12:06 AM
WannaBe
Re: Suspension homology
Hi xxx9,

1) What we've actually proved is that $\displaystyle \phi (Im \partial _p ) \subseteq Im(\partial _{p+1} S(X)) )$ 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 $\displaystyle Im \partial _ 1$ ?

2) After we proved boundaries are copied into boundaries, shouldn't we prove it for cycles as well ? And then my $\displaystyle \phi _{\ast}$ 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!
• Dec 17th 2011, 06:21 AM
xxp9
Re: Suspension homology
Actually what we have shown is $\displaystyle \phi(\partial)=-\partial(\phi)$ for any $\displaystyle p \ge 0$, using this you can easily verify your 1) and 2), by just unwinding the definitions.
• Dec 17th 2011, 07:49 AM
WannaBe
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 $\displaystyle \phi (\partial ) = - \partial ( \phi )$ we can deduce that $\displaystyle \phi (B_p(X)) \subseteq B_{p+1} (S(X))$ . Can you please explain why this is what we needed in order to prove $\displaystyle \phi _{\ast}$ ?

I really need your help in this

• Dec 17th 2011, 08:13 AM
xxp9
Re: Suspension homology
How do you define $\displaystyle \phi(0)$?
• Dec 17th 2011, 08:51 AM
WannaBe
Re: Suspension homology
$\displaystyle w_0 - w_1$ ?
• Dec 17th 2011, 11:22 AM
WannaBe
Re: Suspension homology
NVM, I got it !

THanks a lot!