# Thread: Isomorphic Chain Complexes Question on Homology

1. ## Isomorphic Chain Complexes Question on Homology

Hi,

How do we prove isomorphic complexes have the same cohomology?

If C. and D. are isomorphic, how do we prove

$\displaystyle \mbox{H}_{n}\mbox{(C.)}\cong\mbox{H}_{n}\mbox{(D.) } \ \forall\mbox{n}$

so we need an isomorphism for:

$\displaystyle \mbox{kerd}_{n}\mbox{/}\mbox{imd}_{n+1}\cong \mbox{kerd^\prime}_{n}{/}\mbox{imd}^\prime_{n+1}$

where $\displaystyle \mbox{d}_{n}$, $\displaystyle \mbox{d^\prime}_{n}$ are the differentiations for C. and D. respectively.

I think we must use the individual isomorphisms $\displaystyle \mbox{g:}\mbox{C}_{n}\rightarrow\mbox{D}_{n}$ to find some property/relation between the kernels in each of the exact sequences, and same for the images. I found such properties, just can't see how to finish by constructing the isomorphism!

2. Originally Posted by MaximalIdeal
Hi,

How do we prove isomorphic complexes have the same cohomology?

If C. and D. are isomorphic, how do we prove

$\displaystyle \mbox{H}_{n}\mbox{(C.)}\cong\mbox{H}_{n}\mbox{(D.) } \ \forall\mbox{n}$

so we need an isomorphism for:

$\displaystyle \mbox{kerd}_{n}\mbox{/}\mbox{imd}_{n+1}\cong \mbox{kerd^\prime}_{n}{/}\mbox{imd}^\prime_{n+1}$

where $\displaystyle \mbox{d}_{n}$, $\displaystyle \mbox{d^\prime}_{n}$ are the differentiations for C. and D. respectively.

I think we must use the individual isomorphisms $\displaystyle \mbox{g:}\mbox{C}_{n}\rightarrow\mbox{D}_{n}$ to find some property/relation between the kernels in each of the exact sequences, and same for the images. I found such properties, just can't see how to finish by constructing the isomorphism!

Well, I don't find any relevance in the cohomology in your question. The notation for n-th homology groups is $\displaystyle H_n(C)$, while the notation for n-th cohomology group is $\displaystyle H^n(C)$.

"If an isomorphic chain map is given between two chain complexes, how do we prove that homology groups of two chain complexes are isomorphic?"

If so, define an isomorphic chain map as $\displaystyle f_{\#}$, and first use commutative diagram to show $\displaystyle f_{\#} \partial = \partial f_{\#}$, where $\displaystyle \partial$ is a boundary map. Then, deduce that homology groups of two chain complexes are isomorphic.

3. Yes, it should read homology rather than cohomology, sorry for the confusion.

Thanks for your response, I'm not sure how to use a boundary map as you suggest, is there a way to do it without using boundary maps? or could you explain a little more? Thanks again

4. Originally Posted by MaximalIdeal
Yes, it should read homology rather than cohomology, sorry for the confusion.

Thanks for your response, I'm not sure how to use a boundary map as you suggest, is there a way to do it without using boundary maps? or could you explain a little more? Thanks again
Your $\displaystyle d_n$ and $\displaystyle d_n^\prime$ are probably referring to boundary maps.
If $\displaystyle f_{\#}$ is isomorphic, then $\displaystyle d_n$ and $\displaystyle d_n^\prime$ are basically equivalent having the same kernel and image, since $\displaystyle f_{\#}d_n^\prime=d_nf_{\#}$. Therefore, the above two chain complexes have the same homology groups.

5. Originally Posted by TheArtofSymmetry
Your $\displaystyle d_n$ and $\displaystyle d_n^\prime$ are probably referring to boundary maps.
If $\displaystyle f_{\#}$ is isomorphic, then $\displaystyle d_n$ and $\displaystyle d_n^\prime$ are basically equivalent having the same kernel and image, since $\displaystyle f_{\#}d_n^\prime=d_nf_{\#}$. Therefore, the above two chain complexes have the same homology groups.
The $\displaystyle d_n$ are the differentiations for C., so

$\displaystyle d_{n+1}:C_{n+1}\rightarrow\mbox{C}_{n}$ etc. so that $\displaystyle d_{n+1}d_{n}=0$ for all n.

That makes sense though, just was unsure of the terminology as I'm using Rotman's book and he doesn't seem to always follow standard. I understand the crux of the problem but found it difficult to express! Thanks again