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# 数学代写|代数拓扑作业代写algebraic topology代考|Products

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## 数学代写|代数拓扑作业代写algebraic topology代考|Tensor products of chain complexes and the algebraic K¨unneth theorem

We begin with a discussion about tensor products and Hom of graded $R$ modules.
Definition 3.1.

1. A graded $R$-module $A_{*}$ can be thought of either as a collection of $R$-modules $\left{A_{k}\right}_{k \in \mathbf{Z}}$ or as a module $A=\bigoplus_{k} A_{k}$ with a direct sum decomposition.
2. A homomorphism of graded $R$-modules is an element of the product $\prod_{k} \operatorname{Hom}\left(A_{k}, B_{k}\right)$.
3. The tensor product of graded $R$-modules $A_{}$ and $B_{}$ is the graded $R$-module
$$\left(A_{} \otimes B_{}\right){n}=\bigoplus{p+q=n}\left(A_{p} \otimes B_{q}\right) .$$
4. Define $\operatorname{Hom}\left(A_{}, B_{}\right)$ to be the graded $R$-module, with
$$\operatorname{Hom}\left(A_{}, B_{}\right){n}=\prod{k} \operatorname{Hom}\left(A_{k}, B_{k+n}\right) \text {. }$$
The functors $-\otimes B_{}$ and $\operatorname{Hom}\left(B_{},-\right)$ are adjoint functors from the category of graded $R$-modules to itself.
5. A graded ring is a graded abelian group $R_{}$ together with a (degree zero) map $$R_{} \otimes R_{} \rightarrow R_{}$$
which is associative in the sense that $(a b) c=a(b c)$ where we write $a b$ for the image of $a \otimes b$. Alternatively, a graded ring can be thought of as a ring $R$ with a direct sum decomposition $R=\bigoplus R_{k}$, satisfying $R_{k} \cdot R_{l} \subset R_{k+l}$.
6. A graded ring is commutative if
$$a b=(-1)^{|a||b|} b a,$$
where $a \in R_{|a|}$ and $b \in R_{|b|}$.
7. A graded module $M_{}$ over a graded ring $R_{}$ is a module over $R$ satisfying $R_{k} \cdot M_{l} \subset M_{k+l}$.

We apply these constructions to chain complexes $\left(C_{}, \partial\right)$ and $\left(C_{}^{\prime}, \partial_{*}^{\prime}\right)$. We allow $C_{q}$ and $C_{q}^{\prime}$ to be nonzero for any $q \in \mathbf{Z}$.

## 数学代写|代数拓扑作业代写algebraic topology代考|The Eilenberg-Zilber maps

Recall the statement of the Eilenberg-Zilber theorem. Until further notice, homology and cohomology with coefficients in a ring $R$ are understood, and we omit writing ” $; R “$.

Theorem $3.4$ (Eilenberg-Zilber theorem). Let TOP be the category whose objects are pairs of spaces $(X, Y)$ (we do not assume $Y \subset X$ ) and whose morphisms are pairs $\left(f: X^{\prime} \rightarrow X, g: Y^{\prime} \rightarrow Y\right)$ of continuous maps. Then the two functors
$$F:(X, Y) \mapsto S_{}(X \times Y)$$ and $$F^{\prime}:(X, Y) \mapsto S_{}(X) \otimes S_{}(Y)$$ from TOP 2 to the category of chain complexes are naturally equivalent; i.e. there exist natural transformations $A: F \rightarrow F^{\prime}$ and $B: F^{\prime} \rightarrow F$ so that for any pair $(X, Y)$ the composites $$S_{}(X \times Y) \stackrel{A}{\rightarrow} S_{}(X) \otimes S_{}(Y) \stackrel{B}{\rightarrow} S_{}(X \times Y)$$ and $$S_{}(X) \otimes S_{}(Y) \stackrel{B}{\rightarrow} S_{}(X \times Y) \stackrel{A}{\rightarrow} S_{}(X) \otimes S_{}(Y)$$
are chain homotopic to the identity. Moreover, any two choices of $A$ (resp.
B) are naturally chain homotopic.
In particular, there exist natural isomorphisms
$$H_{n}(X \times Y) \rightarrow H_{n}\left(S_{}(X) \otimes S_{}(Y)\right)$$
for each $n$.

## 数学代写|代数拓扑作业代写ALGEBRAIC TOPOLOGY代考|Cross and cup products

The homology cross product and the Künneth formula. Exercise 32 implies that the natural map
$$\times_{\text {alg }}: H_{p} X \otimes H_{q} Y \rightarrow H_{p+q}\left(S_{} X \otimes S_{} Y\right)$$
given on the chain level by $[a] \otimes[b] \mapsto[a \otimes b]$ is well defined. Denote by $B_{}$ the isomorphism induced by the Eilenberg-Zilber map on homology, so $$B_{}: H_{}\left(S_{}(X) \otimes S_{}(Y)\right) \rightarrow H_{}\left(S_{}(X \times Y)\right)=H_{}(X \times Y) \text {. }$$
Composing $\times_{\text {alg }}$ with $B_{*}$, we obtain
$$\times: H_{p} X \otimes H_{q} Y \rightarrow H_{p+q}(X \times Y) .$$

## 数学代写|代数拓扑作业代写ALGEBRAIC TOPOLOGY代考|TENSOR PRODUCTS OF CHAIN COMPLEXES AND THE ALGEBRAIC K¨UNNETH THEOREM

1. A graded $R$-module $A_{*}$ can be thought of either as a collection of $R$-modules $\left{A_{k}\right}_{k \in \mathbf{Z}}$ or as a module $A=\bigoplus_{k} A_{k}$ with a direct sum decomposition.
2. A homomorphism of graded $R$-modules is an element of the product $\prod_{k} \operatorname{Hom}\left(A_{k}, B_{k}\right)$.
3. The tensor product of graded $R$-modules $A_{}$ and $B_{}$ is the graded $R$-module
$$\left(A_{} \otimes B_{}\right){n}=\bigoplus{p+q=n}\left(A_{p} \otimes B_{q}\right) .$$
4. Define $\operatorname{Hom}\left(A_{}, B_{}\right)$ to be the graded $R$-module, with
$$\operatorname{Hom}\left(A_{}, B_{}\right){n}=\prod{k} \operatorname{Hom}\left(A_{k}, B_{k+n}\right) \text {. }$$
The functors $-\otimes B_{}$ and $\operatorname{Hom}\left(B_{},-\right)$ are adjoint functors from the category of graded $R$-modules to itself.
5. A graded ring is a graded abelian group $R_{}$ together with a (degree zero) map $$R_{} \otimes R_{} \rightarrow R_{}$$
which is associative in the sense that $(a b) c=a(b c)$ where we write $a b$ for the image of $a \otimes b$. Alternatively, a graded ring can be thought of as a ring $R$ with a direct sum decomposition $R=\bigoplus R_{k}$, satisfying $R_{k} \cdot R_{l} \subset R_{k+l}$.
6. A graded ring is commutative if
$$a b=(-1)^{|a||b|} b a,$$
where $a \in R_{|a|}$ and $b \in R_{|b|}$.
7. A graded module $M_{}$ over a graded ring $R_{}$ is a module over $R$ satisfying $R_{k} \cdot M_{l} \subset M_{k+l}$.

## 数学代写|代数拓扑作业代写ALGEBRAIC TOPOLOGY代考|THE EILENBERG-ZILBER MAPS

$$F:(X, Y) \mapsto S_{}(X \times Y)$$ and $$F^{\prime}:(X, Y) \mapsto S_{}(X) \otimes S_{}(Y)$$ from TOP 2 to the category of chain complexes are naturally equivalent; i.e. there exist natural transformations $A: F \rightarrow F^{\prime}$ and $B: F^{\prime} \rightarrow F$ so that for any pair $(X, Y)$ the composites $$S_{}(X \times Y) \stackrel{A}{\rightarrow} S_{}(X) \otimes S_{}(Y) \stackrel{B}{\rightarrow} S_{}(X \times Y)$$ and $$S_{}(X) \otimes S_{}(Y) \stackrel{B}{\rightarrow} S_{}(X \times Y) \stackrel{A}{\rightarrow} S_{}(X) \otimes S_{}(Y)$$
are chain homotopic to the identity. Moreover, any two choices of $A$ (resp.
B) are naturally chain homotopic.
In particular, there exist natural isomorphisms
$$H_{n}(X \times Y) \rightarrow H_{n}\left(S_{}(X) \otimes S_{}(Y)\right)$$

## 数学代写|代数拓扑作业代写ALGEBRAIC TOPOLOGY代考|CROSS AND CUP PRODUCTS

$$\times_{\text {alg }}: H_{p} X \otimes H_{q} Y \rightarrow H_{p+q}\left(S_{} X \otimes S_{} Y\right)$$
given on the chain level by $[a] \otimes[b] \mapsto[a \otimes b]$ is well defined. Denote by $B_{}$ the isomorphism induced by the Eilenberg-Zilber map on homology, so $$B_{}: H_{}\left(S_{}(X) \otimes S_{}(Y)\right) \rightarrow H_{}\left(S_{}(X \times Y)\right)=H_{}(X \times Y) \text {. }$$
Composing $\times_{\text {alg }}$ with $B_{*}$, we obtain
$$\times: H_{p} X \otimes H_{q} Y \rightarrow H_{p+q}(X \times Y) .$$

## Matlab代写

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