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6

From the definitions, there's no reason to expect projective/injective objects of a subcategory to be projective/injective in the ambient category. If $P, B \in \mathcal{A}$, then any map $P \to B$ factors through any epi $A \twoheadrightarrow B$ with $A \in \mathcal{A}$, but if $A, B$ are not in $\mathcal{A}$, there is no reason to expect a lift. For a ...


6

I will answer your question in what follows, but first it might be useful to say something about what derived functors are and what the point is. The idea of derived functors is that if you have a functor which preserves (say) left exactness when applied to a short exact sequence, but not right exactness (e.g. $Hom(X, \text{--})$ for a module $X$ ), then ...


5

Given the hint that the identity morphisms can be weird in ((sets)), we can proceed as follows. Let ((sets)) consist of just the one set $\{0,1\}$ and the one map, from $\{0,1\}$ to itself that sends both 0 and 1 to 0. This one map is then the identity map, composition is as usual, and this identity map is a non-bijective isomorphism.


4

Of course the constant sheaf $\mathbb{Q}$ is not injective in general. Recall that for an injective sheaf $\mathscr{F}$, we have $H^1 ( X, \mathscr{F}) = 0$. However, for sufficiently nice spaces $X$ (say, paracompact), sheaf cohomology with coefficients in a constant sheaf $A$ is isomorphic to singular cohomology with coefficients in $A$; so in particular, ...


3

I'm not sure what exactly you're asking for, but here's a blurb about derived functors that I hope is useful to you. Suppose we have a left-exact functor $F$ on the category $\cal{C}$ of modules over a given commutative ring. (The usual setting is abelian categories, but there's no harm in working with something concrete for a short description. Also, I'm ...


3

Suppose $\def\ZZ{\mathbb Z}0\to\ZZ\xrightarrow{f}E\xrightarrow{g}\ZZ_p\to0$ is a non-split extension. If $E$ has non-zero torsion, then that torsion clearly does not intersect the image of $f$, and therefore is mapped injectively into $\ZZ_p$. You can easily construct, then, a map $\ZZ_p\to E$ which is a section to $g$: this contradicts non-splitness. It ...


3

An object $P$ is projective when, given any morphism $f: P \to A$, and any epimorphism $g: B \to A$, there is a unique morphism $h: P \to B$ such that $g \circ h = f$. Even if an object $P$ is projective in a full subcategory of R-mod, then, it might not be projective in R-mod itself, because in that case the objects $A$ and $B$ might be outside of the ...


3

Because $i_*$ is injective, the images of the boundary maps $\partial$ are zero (by exactness of the sequence given in axiom 4). The exactness of the sequence also means that $j_*$ is surjective. Hence the long exact sequence given by axiom 4 breaks up into short exact sequences $$0\overset{\partial}{\to} ...


3

Recall that if $\Delta^n = [v_0, v_1, \ldots, v_{n-1}, v_n]$ then for a map $\sigma\colon\Delta^n\to X$ we define the boundary operator $\partial\colon C_n(X)\to C_{n-1}(X)$ by $$\partial(\sigma) = \sum_{j=0}^n (-1)^{j} \sigma \mid [v_0,v_1, \ldots, \hat{v}_j,\ldots, v_{n-1},v_n]$$ and if, for an arbitrary chain $\alpha$, we have ...


2

It looks like you're misreading the theorem in a critical detail. It's not the case that any two maps $f_\bullet,g_\bullet:P_\bullet\to P'_\bullet$ are chain homotopic, but that any two such maps induced by the same map $f:X\to Y$ are. (For a counterexample, if $X$ and $Y$ are themselves projective, then two chain maps between $X$ and $Y$ are chain homotopic ...


2

I think it's only a matter of indexing. For example, according to the notation in Weibel's book we would have $\cdots\to \mathscr R_1\to \mathscr R_0\to \mathscr G\to 0$ in the projective resolution, rather than defining $\mathscr R_0 = \mathscr G$. That is, in a projective (or free) resolution, the final map $\mathscr R_1\to\mathscr R_0$ is typically ...


2

This was going to be a comment but it does not quite fit. It sound like your question has something to do with bridging the divide between an intuitive or "picture" understanding of homology and the formal definition of singular or simplicial homology in the language of chain complexes. You can of course define homology without using the language of ...


2

Let $N$ be an $R$-module. We have to show that $N \to N \otimes_R S$ is injective. As $S$ is faithfully flat as an $R$-module, it suffices to prove this after tensoring with $S$. Hence it suffices to show that $N \otimes_R S \to N \otimes_R S \otimes_R S$, $n \otimes s \mapsto n \otimes 1 \otimes s$ is injective. This is true because there is a section, ...


2

What book is it you are quoting? You need to consider the excision property (if you are to follow the given proof), which is given e.g. in Theorem 2.17 and Corollary 2.18 of the following links: http://www.math.toronto.edu/mgualt/MAT1300/Week%206-9%20Term%202.pdf The idea is to use the 5-lemma to show the desired isomorphism. Alternatively, one may show ...


2

The $f$ and $g$ appearing in the proof are not the same as the $f$ and $g$ appearing in the axiom. We have $g\circ f\sim\mbox{Id}_{X,A}\colon (X,A)\to (X,A)$ and so axiom $5$ implies that $(g\circ f)_*=(\mbox{Id}_{X,A})_*$, the other axioms then give us that $(g\circ f)_*=g_*\circ f_*$ by funtoriality, and $(\mbox{Id}_{X,A})_*=\mbox{Id}_{H_k(X,A)}$ also by ...


1

From the first definition we have that $X=\stackrel{\circ}{X-U}\cup \stackrel{\circ}{A}$ because $\overline{U}\subset \stackrel{\circ}{A}$ and $\stackrel{\circ}{X-U}=X-\overline{U}$ then if i put $B=X-U$ i obtain that $H_n(B,B\cap A)=H_n(X-U,A-U)$ where $A$ is the same and $X-U=B$ in this case $X=\stackrel{\circ}{A}\cup \stackrel{\circ}{B}$ so the two ...


1

See the Theorem $1$ and $2$ here, where there is equivalent.


1

If we speak about unital rings and the definition of a module implies $1\cdot m = m$, then there is an answer for an important case where $R$ is a field. In this case, modules are vector spaces. $M\otimes_F N$ consists of elementary tensors if and only if at least one of $M$, $N$ has dimension not greater than one. But it might be obvious for the ...


1

Actually everything you wrote is correct, except the last equation! Why do you say $(u_{n-1} \circ d_n)(v) = 0$? In reality, $d_n v$ is an element of $C_{n-1}$ that you need to write as a sum $v_1' + v_2' + v_3'$ where $v_1' \in \operatorname{im}d_n$ etc. But you know that such a sum is unique, and you already have the decomposition $d_n v = d_n v + 0 + 0$. ...


1

I believe that this is what we need to examine. Although we intuitively want to assume that the identity morphism is the usual "identity" map, this is actually not stated as necessary in the problem. Let $\mathcal{C}$ be a category consisting of a single set $A=\{1,2\}$ as its object, morphisms as set maps, and composition as usual. Let $Mor(A,A)$ consist ...


1

The trick to do (1) goes like this. If you have $x \in H_k(X)$, you apply $ir$ to get $ir(x) \in im i_*$. The cool thing is than you can then just check that $x - irx$ is in the kernel of $r$, and so $x = irx + (x - irx)$ is your decomposition. To show it's direct, you take $0 = iy + z$. Then applying $r$, we get $0 = riy = y$, and so the sum is direct. The ...


1

This is an application of a result of homological algebra which is: If $0\to N\stackrel{i}{\to} M\stackrel{j}\to P\to 0$ is an exact sequence and $r:M\to N$ such that $r\circ i=id_{N}$ then $\ker r\simeq P$ and $\text{Im}(i)\simeq N$, and $M=\text{Im}(i)\oplus \ker r \simeq N\oplus P$. Proof: Because $i$ is injective we have $\text{Im}(i)\simeq N$. For ...


1

For every $i$ we have $C_i=B_i/A$. Taking direct limit, when $B'$ denotes the limit of $(B_i)$, we get $$C=B'/A,$$ and $B'$ is a submodule of $B$. But we know that $C=B/A$, hence $B'=B$.


1

@Vrouvrou: You have asked many questions on the subject of Morse theory. I have the feeling any book on the subject must discuss this on the first page. The basic observation is (under suitable compactness assumptions) the following. Let $f:M\rightarrow \mathbb{R}$ be a smooth function. And $c-\epsilon$ and $c+\epsilon$ are regular values such that the ...


1

If a complex is exact, its homology is zero. It follows that if its homology is non-zero, the complex is not exact and, intuitively, the larger the homology, the farther the complex is from being exact. There is nothing more than this.



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