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I met so many problems when I study homological algebra by myself. Thus, I really would like to see the answers. Hopefully, everyone can help me (my big thanks).

1) When we create torsion functor from a resolution of M illustrated by:

$...\overset{d_{3}}{\rightarrow} X_{2} \overset{d_{2}}{\rightarrow} X_{1} \overset{d_{1}}{\rightarrow} X_{0} \overset{\epsilon}{\rightarrow}M \rightarrow 0$

We eliminate M in above resolution then we have:

$...\overset{d*_{3}}{\rightarrow} X_{2} \overset{d*_{2}}{\rightarrow} X_{1} \overset{d*_{1}}{\rightarrow} X_{0} \rightarrow 0$

Now the above sequence is semi-exact. After taking homology from the product tensor of it and module B, we have torsion functor. I would like to ask that $d_{i}$ and $d*_{i}$ is the same with $i\ge 1$ or not? If it is the same (I mean that when we take the homology), $Tor_{n}(M,B)$ will be equal zero with $n\ge 1$ because $Imd*_{i+1}=Imd_{i+1}=Kerd_{i}=Kerd*_{i}$ then $Im(d*_{i+1}\otimes 1) = Ker(d*_{i}\otimes 1)$

2) We have cohomology by effecting functor $Hom(-,B)$ to semi-exact sequence and then take homology. I would like to ask that it will be also called cohomology when we use functor $Hom(B,-)$ or not?

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    $\text{Im} d^_{i+1} = \text{Ker}d^{i}$ does not imply $\text{Im} (d^*{i+1}\otimes 1) = \text{Ker}(d^*_{i}\otimes 1)$. Using $\text{Hom}(B,-)$ will give Ext groups if you apply it to an injective resolution, but I don't think you get anything particularly useful if you apply it to a projective resolution. – Rolf Hoyer May 24 '16 at 05:19
  • Sorry! The first one, from $Imd_{i+1}=Kerd{i}$ we do not have $Im(d*{i+1}\otimes 1)=Ker(d*_{i}\otimes 1)$. Could you explain it more than that? I think that it is equal. The second one, I just wanna ask that it sill be called cohomology or not? I mean it is cohomology or homology because when we take the $Hom(B,-)$ the index of semi-exact sequence is not reserved so we do not call it being cohomology, right? Just homology – Thế Long Lê May 24 '16 at 05:28
  • Ok I got it. Thank you so much – Thế Long Lê May 24 '16 at 08:02

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I think the name homology or cohomology functor depends on the used functor invariant or covariant functor .is it?

m.idaya
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