Does the canonical map of the universal coefficient theorem induce an isomorphism?












2












$begingroup$


Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence:
$$
0 rightarrow text{Ext}^1_R(H(C), R) rightarrow H(C^*) overset{h}{rightarrow} text{Hom}_R(H(C), R) rightarrow 0.
$$
My question is, is the following homomorphism
$$ bar{h}: H(C^*)/text{(tor)} rightarrow text{Hom}_R(H(C), R)$$
an isomorphism?



A torsion $[f] in H(C^*)$ is annihilated by $h$, since if there is an $a in R$ such that $a[f] = 0$, then $$ah([f])([z]) = aleft< [f], [z]right> = left< a[f], [z]right> = 0,$$ so $h([f]) = 0$ and the map is well-defined. On the other hand, from the structure theorem, we may write $$H(C) cong R^r oplus_i R/(a_i)$$ and we have $$text{Hom}_R(H(C), R) cong text{Hom}_R(R^r, R) oplus_i text{Hom}_R(R/(a_i), R) cong R^r$$ and $$text{Ext}_R(H(C), R) cong text{Ext}_R(R^r, R) oplus_i text{Ext}_R(R/(a_i), R) cong bigoplus_i R/(a_i)$$ so we have $$H(C^*) cong R^r oplus_i R/(a_i).$$



So we can say that $text{Ext}_R(H(C), R)$ is (abstractly) isomorphic to the torsion submodule of $H(C^*)$, but not canonically as discussed in this question.



Thanks in advance.






homological-algebra







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$endgroup$












  • $begingroup$
    Maybe it is helpful to know that the short exact sequence splits?
    $endgroup$
    – Johannes Huisman
    Dec 19 '18 at 14:48
















2












$begingroup$


Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence:
$$
0 rightarrow text{Ext}^1_R(H(C), R) rightarrow H(C^*) overset{h}{rightarrow} text{Hom}_R(H(C), R) rightarrow 0.
$$
My question is, is the following homomorphism
$$ bar{h}: H(C^*)/text{(tor)} rightarrow text{Hom}_R(H(C), R)$$
an isomorphism?



A torsion $[f] in H(C^*)$ is annihilated by $h$, since if there is an $a in R$ such that $a[f] = 0$, then $$ah([f])([z]) = aleft< [f], [z]right> = left< a[f], [z]right> = 0,$$ so $h([f]) = 0$ and the map is well-defined. On the other hand, from the structure theorem, we may write $$H(C) cong R^r oplus_i R/(a_i)$$ and we have $$text{Hom}_R(H(C), R) cong text{Hom}_R(R^r, R) oplus_i text{Hom}_R(R/(a_i), R) cong R^r$$ and $$text{Ext}_R(H(C), R) cong text{Ext}_R(R^r, R) oplus_i text{Ext}_R(R/(a_i), R) cong bigoplus_i R/(a_i)$$ so we have $$H(C^*) cong R^r oplus_i R/(a_i).$$



So we can say that $text{Ext}_R(H(C), R)$ is (abstractly) isomorphic to the torsion submodule of $H(C^*)$, but not canonically as discussed in this question.



Thanks in advance.






homological-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe it is helpful to know that the short exact sequence splits?
    $endgroup$
    – Johannes Huisman
    Dec 19 '18 at 14:48














2












2








2





$begingroup$


Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence:
$$
0 rightarrow text{Ext}^1_R(H(C), R) rightarrow H(C^*) overset{h}{rightarrow} text{Hom}_R(H(C), R) rightarrow 0.
$$
My question is, is the following homomorphism
$$ bar{h}: H(C^*)/text{(tor)} rightarrow text{Hom}_R(H(C), R)$$
an isomorphism?



A torsion $[f] in H(C^*)$ is annihilated by $h$, since if there is an $a in R$ such that $a[f] = 0$, then $$ah([f])([z]) = aleft< [f], [z]right> = left< a[f], [z]right> = 0,$$ so $h([f]) = 0$ and the map is well-defined. On the other hand, from the structure theorem, we may write $$H(C) cong R^r oplus_i R/(a_i)$$ and we have $$text{Hom}_R(H(C), R) cong text{Hom}_R(R^r, R) oplus_i text{Hom}_R(R/(a_i), R) cong R^r$$ and $$text{Ext}_R(H(C), R) cong text{Ext}_R(R^r, R) oplus_i text{Ext}_R(R/(a_i), R) cong bigoplus_i R/(a_i)$$ so we have $$H(C^*) cong R^r oplus_i R/(a_i).$$



So we can say that $text{Ext}_R(H(C), R)$ is (abstractly) isomorphic to the torsion submodule of $H(C^*)$, but not canonically as discussed in this question.



Thanks in advance.






homological-algebra







share|cite|improve this question











$endgroup$




Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence:
$$
0 rightarrow text{Ext}^1_R(H(C), R) rightarrow H(C^*) overset{h}{rightarrow} text{Hom}_R(H(C), R) rightarrow 0.
$$
My question is, is the following homomorphism
$$ bar{h}: H(C^*)/text{(tor)} rightarrow text{Hom}_R(H(C), R)$$
an isomorphism?



A torsion $[f] in H(C^*)$ is annihilated by $h$, since if there is an $a in R$ such that $a[f] = 0$, then $$ah([f])([z]) = aleft< [f], [z]right> = left< a[f], [z]right> = 0,$$ so $h([f]) = 0$ and the map is well-defined. On the other hand, from the structure theorem, we may write $$H(C) cong R^r oplus_i R/(a_i)$$ and we have $$text{Hom}_R(H(C), R) cong text{Hom}_R(R^r, R) oplus_i text{Hom}_R(R/(a_i), R) cong R^r$$ and $$text{Ext}_R(H(C), R) cong text{Ext}_R(R^r, R) oplus_i text{Ext}_R(R/(a_i), R) cong bigoplus_i R/(a_i)$$ so we have $$H(C^*) cong R^r oplus_i R/(a_i).$$



So we can say that $text{Ext}_R(H(C), R)$ is (abstractly) isomorphic to the torsion submodule of $H(C^*)$, but not canonically as discussed in this question.



Thanks in advance.






homological-algebra




homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 6:47







Taketo Sano

















asked Dec 19 '18 at 5:33









Taketo SanoTaketo Sano

1134




1134












  • $begingroup$
    Maybe it is helpful to know that the short exact sequence splits?
    $endgroup$
    – Johannes Huisman
    Dec 19 '18 at 14:48


















  • $begingroup$
    Maybe it is helpful to know that the short exact sequence splits?
    $endgroup$
    – Johannes Huisman
    Dec 19 '18 at 14:48
















$begingroup$
Maybe it is helpful to know that the short exact sequence splits?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:48




$begingroup$
Maybe it is helpful to know that the short exact sequence splits?
$endgroup$
– Johannes Huisman
Dec 19 '18 at 14:48










1 Answer
1






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oldest

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$begingroup$

It appears to me that the image of $newcommandExt{operatorname{Ext}}Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $Ext$, and yet also every element of the image of $Ext$ lies in the torsion subgroup. Thus the image of $Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.



There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.






share|cite|improve this answer









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    $begingroup$

    It appears to me that the image of $newcommandExt{operatorname{Ext}}Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $Ext$, and yet also every element of the image of $Ext$ lies in the torsion subgroup. Thus the image of $Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.



    There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It appears to me that the image of $newcommandExt{operatorname{Ext}}Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $Ext$, and yet also every element of the image of $Ext$ lies in the torsion subgroup. Thus the image of $Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.



      There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It appears to me that the image of $newcommandExt{operatorname{Ext}}Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $Ext$, and yet also every element of the image of $Ext$ lies in the torsion subgroup. Thus the image of $Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.



        There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.






        share|cite|improve this answer









        $endgroup$



        It appears to me that the image of $newcommandExt{operatorname{Ext}}Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $Ext$, and yet also every element of the image of $Ext$ lies in the torsion subgroup. Thus the image of $Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.



        There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 21:37









        jgonjgon

        15.6k32143




        15.6k32143






























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