# 线性代数网课代修|同调代数代写homological algebra代考|MATH812

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

There is a remarkable relationship between Hom and $\otimes$. The key idea is that a function of two variables, say, $f: A \times B \rightarrow C$, can be viewed as a oneparameter family of functions of one variable: if we fix $a \in A$, then define $f_{a}: B \rightarrow C$ by $b \mapsto f(a, b)$. Recall Proposition 2.51: if $R$ and $S$ are rings and $A_{R}$ and $R_{R} B_{S}$ are modules, then $A \otimes_{R} B$ is a right $S$-module, where $(a \otimes b) s=a \otimes(b s)$. Furthermore, if $C_{S}$ is a module, then $\operatorname{Hom}{S}(B, C)$ is a right $R$-module, where $(f r)(b)=f(r b)$; thus $\operatorname{Hom}{R}\left(A, \operatorname{Hom}{S}(B, C)\right)$ makes sense, for it consists of $R$-maps between right $R$-modules. Finally, if $F \in \operatorname{Hom}{R}\left(A, \operatorname{Hom}{S}(B, C)\right)$, we denote its value on $a \in A$ by $F{a}$, so that $F_{a}: B \rightarrow C$, defined by $F_{a}: b \mapsto F(a)(b)$, is a one-parameter family of functions. There are two versions of the adjoint isomorphism, arising from two ways in which $B$ can be a bimodule (either ${ }{R} B{S}$ or ${ }{S} B{R}$ ).

Remark. In more detail, fixing any two of $A, B, C$, each $\tau_{A, B, C}$ is a natural isomorphism:
$\operatorname{Hom}{S}\left(\square \otimes{R} B, C\right) \rightarrow \operatorname{Hom}{R}\left(\square, \operatorname{Hom}{S}(B, C)\right)$,
$\operatorname{Hom}{S}\left(A \otimes{R} \square, C\right) \rightarrow \operatorname{Hom}{R}\left(A, \operatorname{Hom}{S}(\square, C)\right)$,
$\operatorname{Hom}{S}\left(A \otimes{R} B, \square\right) \rightarrow \operatorname{Hom}{R}\left(A, \operatorname{Hom}{S}(B, \square)\right)$.
For example, if $f: A \rightarrow A^{\prime}$, there is a commutative diagram
Proof. To prove that $\tau=\tau_{A, B, C}$ is a $\mathbb{Z}$-map, let $f, g: A \otimes_{R} B \rightarrow C$. The definition of $f+g$ gives, for all $a \in A$,
\begin{aligned} \tau(f+g){a}: b & \mapsto(f+g)(a \otimes b) \ &=f(a \otimes b)+g(a \otimes b) \ &=\tau(f){a}(b)+\tau(g)_{a}(b) \end{aligned}

## 线性代数作业代写linear algebra代考|Projective Modules

The functors $\operatorname{Hom}{R}(X, \square)$ and $\operatorname{Hom}{R}(\square, Y)$ almost preserve short exact sequences; they are left exact functors. Similarly, the functors $\square \otimes_{R} Y$ and $X \otimes R \square$ almost preserve short exact sequences; they are right exact functors. Are there any functors that do preserve short exact sequences?

Definition. A covariant functor $T: R \mathbf{M o d} \rightarrow \mathbf{A b}$ is an exact functor if, for every exact sequence
$$0 \rightarrow A \stackrel{i}{\rightarrow} B \stackrel{p}{\rightarrow} C \rightarrow 0,$$
the sequence
$$0 \rightarrow T(A) \stackrel{T(i)}{\longrightarrow} T(B) \stackrel{T(p)}{\longrightarrow} T(C) \rightarrow 0$$
is also exact. A contravariant functor $T:{ }_{R} \mathbf{M o d} \rightarrow \mathbf{A b}$ is an exact functor if there is always exactness of
$$0 \rightarrow T(C) \stackrel{T(p)}{\longrightarrow} T(\bar{B}) \stackrel{T(i)}{\longrightarrow} T(A) \rightarrow 0 .$$
In Theorem 2.35, we saw that every left module is a quotient of a free left module. Here is a property of free modules that does not mention bases.

$\operatorname{Hom} S(\square \otimes R B, C) \rightarrow \operatorname{Hom} R(\square, \operatorname{Hom} S(B, C)$ ),
$\operatorname{Hom} S(A \otimes R \square, C) \rightarrow \operatorname{Hom} R(A, \operatorname{Hom} S(\square, C))$,
$\operatorname{Hom} S(A \otimes R B, \square) \rightarrow \operatorname{Hom} R(A, \operatorname{Hom} S(B, \square))$.

$$\tau(f+g) a: b \mapsto(f+g)(a \otimes b) \quad=f(a \otimes b)+g(a \otimes b)=\tau(f) a(b)+\tau(g)_{a}(b)$$

## 线性代数作业代写linear algebra代考|Projective Modules

$$0 \rightarrow A \stackrel{i}{\rightarrow} B \stackrel{p}{\rightarrow} C \rightarrow 0,$$

$$0 \rightarrow T(A) \stackrel{T(i)}{\longrightarrow} T(B) \stackrel{T(p)}{\longrightarrow} T(C) \rightarrow 0$$

$$0 \rightarrow T(C) \stackrel{T(p)}{\longrightarrow} T(\bar{B}) \stackrel{T(i)}{\longrightarrow} T(A) \rightarrow 0 .$$

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions