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6

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 ...


5

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 ...


4

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 ...


4

$\text{Ext}^1(-, -)$ sends direct sums in the first variable to direct products, and as mentioned in the comments, as an abstract abelian group $$S^1 \cong \left( \bigoplus_X \mathbb{Q} \right) \oplus \mathbb{Q}/\mathbb{Z}$$ where $X$ is uncountable. So it suffices to compute $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ and $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, ...


2

As requested by Julian Kuelshammer, I shall put my comment here so that the question can be marked as answered. If we have two short exact sequences of modules over a ring $R$, say $0\to X\to Y\to Z\to 0$ and $0\to X'\to Y'\to Z'\to 0$, then it does not hold in general that $X\cong X'$ and $Z\cong Z'$ imply that the two short exact sequences are equivalent, ...


2

Whether these chain complexes are literally equal depends on the precise set-theoretic definitions you have chosen for all the notation involved. For instance, a common definition of ${\bigoplus}_{i \in I} C^{sing}_n(X_i;R)$ is the set of all functions $f$ with domain $I$ such that $f(i)\in C^{sing}_n(X_i;R)$ for each $i\in I$ and $f(i)$ is the zero element ...


2

The component $D^2$ is contractible, so $H_*(D^2\times S^1)=H_*(S^1)$, because homology are homotopy invariant. It is well-known that $H_1(S^1)=\mathbb Z$.


2

Since $\mathfrak{add}\ _AA = \mathfrak{add}\ M$, we have that $A$ and $C=\text{End}_A(M)$ are Morita-equivalent, and that the functor $\text{Hom}_A(M,-): \text{Mod }A \to \text{Mod }C$ is an equivalence of categories. Since equivalences of categories preserve injective objects, the property is proved.


2

Now I have some kind of answer. At first, note that short exact sequence $$0\to H^n(X,\mathbb Z)\otimes R\to H^n(X,R)\to \mathrm{Tor}(H^{n+1}(X,\mathbb Z), R)\to0$$ occurs for finitely-generated (as $\mathbb Z$-module) ring $R$, because of equality $C^*(X,R)=C^*(X,\mathbb Z)\otimes R$. Suggestions about being $X$ a compact or a $CW$-complex are ...


1

Think about the case where $X = \mathbb{R}^2, U = {(0,0)}$ and $V = \mathbb{R}^2 \setminus (0,0)$. We now have that $U \cap V = \emptyset. $ Then $H_1(X) = 0, H_1(U \cap V) = 0$, which by exactness implies $H_1(U) \oplus H_1(V) = 0.$ But, $H_1(U) \oplus H_1(V) = 0 \oplus \mathbb{Z}$ So the sequence cannot be exact.


1

$\newcommand{\Ch}{ C_{\tilde h}} \newcommand{\CP}{\mathbb{CP}^2}$ We will make use of natural inclusions $\alpha_i\colon S^2\hookrightarrow C_{\tilde h}$ and natural projections $\pi_j\colon \Ch\to \CP$. The composition $\pi_j\circ \alpha_j$ is the standard inclusion of $S^2$ in $\mathbb{CP}^2$ and the compositions $\pi_j\circ\alpha_i$ are the constant map ...


1

Weibel's An Introduction to Homological Algebra is a standard text.


1

The easiest way to show this equivalence is slightly indirect: statements (1) and (2) are both equivalent to: (3): $M \simeq N \oplus P$ and under this isomorphism $f: N \rightarrow M$ is the map $\iota: n \mapsto (n, 0)$ and $g: M \rightarrow P$ is the map $\pi: (n, p) \mapsto p$. It's easy to see that with these maps, we get the canonical split exact ...


1

The 'must be faulty' proof (or proofs, with the addendum) above looks (look) correct, and the original exercise, as it appeared in the text, was wrong. In fact - see http://www.math.rutgers.edu/~weibel/Hbook-corrections.html for corrections to Charles Weibel's Introduction to Homological Algebra. The page (currently) has links to the 1994 hard- and 1995 ...


1

I was getting to it, but the addendum convinces me totally. I don't think you will find this exact question in my book Acyclic Models, but you will find a lot of arguments of this sort.


1

Let $P$ be a Euler-Poincaré map. We have the so called "fundamental" short exact sequences $$0\longrightarrow B_n\longrightarrow Z_n \longrightarrow H_n \longrightarrow 0\\ 0 \longrightarrow Z_n \longrightarrow C_n \longrightarrow B_{n+1} \longrightarrow 0$$ so that $P(C_n)=P(Z_n)+P(B_{n+1})$ and $ P(Z_n)=P(B_n)+P(H_n)$ Then $$\begin{align}\chi(C)&= ...


1

Note that $E^k \setminus \{p\}$ deformation retracts to $S^{k-1}$ by $r : E^k \setminus \{p\} \to S^{k-1}, \; x \mapsto \frac x {\| x \|}$. Therefore, $\mathrm{id} \sqcup _f r : X \sqcup _f (E^k \setminus \{ p \}) = V \to X \sqcup _f S^{k-1} \simeq X$ is a retract of $V$ to $X$. To clarify why $X \sqcup _f S^{k-1} \simeq X$ let us show that, in general, if ...


1

There is a more general notion of a mapping cone. That is, an adjunction space corresponding to $f:S^{k-1} \to X$ is the same as a mapping cone $C_f$. For those you find plenty of answers which use the same strategy, hence you can build your intuition and see formulas by considering e.g. https://en.wikipedia.org/wiki/Mapping_cone_(topology), Homology of ...



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