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Math Help - Isomorphic Chain Complexes Question on Homology

  1. #1
    Newbie MaximalIdeal's Avatar
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    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

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

    so we need an isomorphism for:

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

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

    I think we must use the individual isomorphisms \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!

    Thanks in advance
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  2. #2
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    Quote Originally Posted by MaximalIdeal View Post
    Hi,

    How do we prove isomorphic complexes have the same cohomology?

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

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

    so we need an isomorphism for:

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

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

    I think we must use the individual isomorphisms \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!

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

    Is your question saying

    "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 f_{\#}, and first use commutative diagram to show f_{\#} \partial = \partial f_{\#}, where \partial is a boundary map. Then, deduce that homology groups of two chain complexes are isomorphic.
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  3. #3
    Newbie MaximalIdeal's Avatar
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    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
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  4. #4
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    Quote Originally Posted by MaximalIdeal View Post
    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 d_n and d_n^\prime are probably referring to boundary maps.
    If f_{\#} is isomorphic, then d_n and d_n^\prime are basically equivalent having the same kernel and image, since f_{\#}d_n^\prime=d_nf_{\#}. Therefore, the above two chain complexes have the same homology groups.
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  5. #5
    Newbie MaximalIdeal's Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    Your d_n and d_n^\prime are probably referring to boundary maps.
    If f_{\#} is isomorphic, then d_n and d_n^\prime are basically equivalent having the same kernel and image, since f_{\#}d_n^\prime=d_nf_{\#}. Therefore, the above two chain complexes have the same homology groups.
    The d_n are the differentiations for C., so

    d_{n+1}:C_{n+1}\rightarrow\mbox{C}_{n} etc. so that 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
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