8

Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to higher derived functors on other objects. The following illustration may make this feel less surprising. Let $A$ be any object and take a short exact sequence $$0\...


5

The main motivation for wanting $d \circ d = 0$ is because it arises naturally in so many applications. As others have already said in comments, exact sequences are common and important, and you often get chain complexes by applying functors to exact sequences. At that point, homology measures how far the new sequence is from being exact: often a very ...


5

You can use the matrix $M=\begin{pmatrix}\cos&-\sin \\ \sin&\cos\end{pmatrix}$. Note that if $f\in P$, then $f\times \sin\in R$, similarly $f\times \cos\in R$. Thus if $\begin{pmatrix}f\\g\end{pmatrix}\in P\oplus P$ then $M\begin{pmatrix}f\\g\end{pmatrix}= \begin{pmatrix}f\times \cos-g\times\sin\\ f\times\sin+g\times\cos\end{pmatrix} \in R\oplus R$. ...


5

You only need projective resolutions; for Ext you resolve the first argument, not the second. If $A$ is an abelian group, first consider any surjection $f : \bigoplus_I \mathbb{Z} \to A$ from a free abelian group to $A$. The kernel of this surjection is another free abelian group $\bigoplus_J \mathbb{Z}$, because subgroups of free abelian groups are free ...


4

This assuming that $\phi([h]^{-1})=-[[h]]\in H_1(X)$, how can I prove this? This is simply because $\phi$ is a group homomorphism. Generally if $f:G\to H$ is a group homomorphism then you have $$e=f(e)=f(gg^{-1})=f(g)f(g^{-1})$$ and so $f(g^{-1})$ is the inverse of $f(g)$. In symbolic notation $f(g)^{-1}=f(g^{-1})$. It's just that $H_1(X)$ is abelian and ...


3

In the commutative case, let $K$ be the fraction field of $R$. Then you have for any module $A$, an exact sequence $0\to A_{\mathrm{tor}}\to A\to A\otimes_R K\to A_1\to 0$, where $A_1$ is defined by the above sequence. Then, by snake lemma, you get an exact sequence, $0\to A_{\mathrm{tor}}\to B_{\mathrm{tor}}\to C_{\mathrm{tor}}\to A_1\to B_1\to C_1\to 0$.


3

This is false. For instance, let $k$ be a field $R=k\oplus V$ where $V$ is an infinite-dimensional $k$-vector space, with multiplication on $R$ defined by $(a,v)\cdot(b,w)=(ab,aw+bv)$. Let $M=R$ and construct $L$ as follows. Start with $L_0=V$, which is an ideal of $R$. For each finite dimensional subspace $W\subset V$ and each $R$-linear homomorphism $f:...


3

You want the first square to be a pushout square. So take $Z=(X\oplus Y)/N$ where $N=\{(f(m),-g(m)):m\in M\}$.


3

What about the bundle $\mathbb{Z}/2 \to S^n \to \mathbb{R}P^n$ where $n>0$ is even? Then for all $k>0$ the groups $H_k(\mathbb{Z}/2)$ and $H_k(\mathbb{R}P^n)$ are all in the Serre class of finite groups, but $H_n(S^n)\cong \mathbb{Z}$.


3

See here for instance to get the definition and main results about the spectral sequence associated to a double complex. Then we have our projective resolution $P_\bullet \to M$, our injectie resolution $N\to I^\bullet$, with which we form a double complex $K^{p,q} = \hom(P_p, I^q)$. One simply has to be a bit careful with the signs, but it's just ...


3

This is literally the definition of $H_1$. By definition, $H_n(X)$ is the quotient of the group of $n$-cycles by the subgroup of boundaries. So since $H_1(point)$ is trivial, that means every $1$-cycle in $point$ is a boundary. In particular, $f$ (which as a constant map can be considered as taking values in $point$) is the boundary of some $2$-chain in $...


2

For finite dimensional algebras, flat modules are projective, so the projective dimension of $M$ is the same as its weak dimension, which is $$\sup\{d\mid \text{Tor}^A_d(M,-)\neq0\}.$$ Since the class of left modules $X$ such that $\text{Tor}^A_d(M,X)=0$ is closed under coproducts and extensions, and every module is an iterated extension of coproducts of ...


2

As @Arnaud D mentioned, we should think of $n$ as fixed. We are applying $Hom_R(Z_n,\square)$ to the complex $C$ and then looking at the $n$th (same $n$) homology of the result. Let the complex $C$ be as follows: $$\cdots\xrightarrow{} C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1}\xrightarrow{} \cdots$$ Apply $Hom_R(Z_n,\square)$ and assume ...


2

I'll use "complex" to mean "nonnegatively graded (cochain) complex". Of course if you have a bounded below cochain complex, up to a shift one may assume we are in this situation. Here's a direct argument, that singles out the essential bits of the "model-category argument" that I sketched in the comments. It turns out that the important things you need are ...


2

The group is isomorphic to $\mathbb{Z} ^2$. Define the homomorphism $f:\mathbb{Z}^3 \rightarrow \mathbb{Z}^2$ by $(a,b,c)\rightarrow(b,2a-c)$. This map is onto and its kernel is $\langle (1,0,2) \rangle$. This is because you are in the kernel precisely when $b=0$ and $a,c$ satisfy $2a=c$, clearly any value of $a$ works and $c$ must be $2a$. You can tell ...


2

Your reasoning is essentially correct. In fact, it's completely correct if your target category has all colimits. (Ignoring some minor typos, and things that aren't really expressed clearly in places.) The only major issue with it is the final natural isomorphism. The statement you want to prove is that If $J:I\to C$ is a diagram in $C$ (i.e. a functor ...


2

I guess it's no harder to do it from scratch: suppose $\left \langle a,\lambda _{i} \right \rangle$ is a colimit cocone on the functor $A$. Then $\left \langle La,L\lambda _{i} \right \rangle$ is a cocone on $LA$. Let $\left \langle x,\mu _{i} \right \rangle$ be any other cocone on $LA$. Then taking adjoints, we get a cocone on $A$, namely, $\left \langle ...


1

The definition of termwise split is merely that for each $n$, the short exact sequence $0\to A^n\to B^n\to C^n\to 0$ splits. This gives an isomorphism $B^n\cong A^n\oplus C^n$, but crucially, this splitting is not assumed to be compatible with the differentials in the chain complex. So, for instance, for each $n$ we have a map $C^n\to B^n$ which splits the ...


1

The key idea is that you can use whichever projective resolution of $M$ you want to compute $(L_nF)(M)$. In particular, when $M$ itself is projective, there's a really easy projective resolution: $$\cdots\to 0\to 0\to 0\to M\stackrel{1}\to M\to 0,$$ with $C_0=M$ and $C_n=0$ for all $n\geq 1$. Applying the definition, we see that $(L_nF)(M)$ is the $n$th ...


1

So Hatcher makes two claims few lines earlier that are important: $\varphi_*$ commutes with homology long exact sequence $\varphi_*$ commutes with induced homomorphisms $f_*$ for any $f:(X,A)\to(Y,B)$ Both are easy to show (hopefuly). What is harder is that I think that 1. implies that $\varphi_*$ also commutes with the Mayer-Vietoris sequence. Which will ...


1

I think I have it (for the first)! For the first: $\text{Tor}_i(M,N)$ is annihilated by $\text{Ann}(M)$ and $\text{Ann}(N)$ because element of $\text{Tor}$ are nothing else classe of elements of $P_i\otimes N$ or $P_i\otimes M$. Here $\text{Tor}_1(R/I,R/J)$ is annihilated by $I$ and $J$ so if $R=I+J$ it is annihilated by 1 and the conclusion come.


1

If $R$ is commutative, the statement is completely elementary --- not homological at all. Suppose $I+J=R$, so that $a+b=1$ for some $a\in I$, $b\in J$; we want to show $I\cap J\subseteq IJ$. Let $x\in I\cap J$. Then $ax\in IJ$ and $xb\in IJ$. So $x=x(a+b)=ax+xb\in IJ$.


1

My favourite proof for (1) uses the vector space duality $D:=\mathrm{Hom}_F(-,F)$. It is easy to see that a finite dimensional $R$-module $X$ is projective (respectively invective) if and only if $D(X)$ is injective (respectively projective). Now consider the map $$ \tau \in D(R), \quad a_0+a_1x+\cdots+a_{n-1}x^{n-1} \mapsto a_{n-1}. $$ Then $\tau$ induces ...


1

The expansion $R\to R$ can be constructed by mapping $1$ to $a_k+a_{k+1}x+\cdots+a_{n-1}x^{n-1-k}$ (and I think k need not divide n). The projective resolution is a series of $R$'s and maps are multiplication of $x$ and $x^{n-1}$ alternatively, ending by $x$ to the ideal. $\require{AMScd}$ \begin{CD} ... @>x^{n-1}>> R@>x>>R@>x^{n-1}...


1

It probably cant get better than the author of the article answering himself but here a small addition that also answers the question and gives a direct interpretation of this Tor, namely it counts something. Since we do homological algebra, we can assume that the algebra is basic. As in Jeremy Rickards answer one has $Tor_d^A(M,A/radA)=DExt_A^d(M,D(A/radA))$...


1

Up to isomorphism, $\operatorname{tot} X$ doesn't change if you switch rows and columns. so whatever is true for columns is true for rows. Alternatively, Theorem 12.5.4 is true, with the same proof, if you replace $H_{II}H_I$ with $H_IH_{II}$.


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