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The Serre spectral sequence is an imperfect tool for calculating the homology of the fiber when the base space is not simply connected. Then you have to use (co)homology with local coefficient systems and you can't get the inductive methods started, because these types of (co)homology do not give you enough information. In your case, it's instructive to ...

6

Long exact sequence of fibration $pt\to T^n$ gives us that $\pi_k(\Omega T^n)=\pi_{k+1}(T^n)$, so $\pi_k(\Omega T^n)=0$ when $k>0$. Since $\pi_1(T^n)=\mathbb Z^n$ transitive acts by permutations on the connected components of $\Omega T^n$, all the components are homeomorphic. And as $T^n$ is a $CW$-complex, the space $\Omega T^n$ has the homotopy type of ...

5

At Cornell, we had recently a class on Homological algebra taught by Yuri Berest. I think you have enough background for reading the notes to that class. You can find them here. I think he did quite a good job of carefully going through the main basic things like abelian and triangulated categories, derived functors et.c. But at the same time, he was trying ...

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It is enough to prove it preserves short exact sequences: $\;0\to M\to N\to P\to 0$. As the tensor product is right-exact, and $S^{-1}M\simeq M\otimes_A S^{-1}A$, it is even enough to prove it preserves injectivity. So consider an injective morphism $\varphi\colon M\to N$ and suppose $\;S^{-1}\varphi\Bigl(\dfrac ms\Bigr)=0$ in $S^{-1}N$. This means there ...

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It simply doesn't follow that $g$ is a morphism in $S$. For example, $A$ might be the category of abelian groups, $S$ might be the full subcategory of finitely generated abelian groups, and the target of $g$ might be an infinitely generated abelian group. But the proof is very easy to repair: just define $g = \text{coker}(f) \in A$ in the first place. Then ...

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