这是一份durham杜伦大学MATH41120-WE01/MATH4161-WE01作业代写的成功案例
Proof: The $\alpha_{i}$ ‘s will be constructed inductively. Since the $F_{i}$ ‘s are free, it suffices to define each $\alpha_{i}$ on a basis for $F_{i}$. To define $\alpha_{0}$, observe that surjectivity of $f_{0}^{\prime}$ implies that for each basis element $x$ of $F_{0}$ there exists $x^{\prime} \in F_{0}^{\prime}$ such that $f_{0}^{\prime}\left(x^{\prime}\right)=\alpha f_{0}(x)$, so we define $\alpha_{0}(x)=x^{\prime}$. We would like to define $\alpha_{1}$ in the same way, sending a basis element $x \in F_{1}$ to an element $x^{\prime} \in F_{1}^{\prime}$ such that $f_{1}^{\prime}\left(x^{\prime}\right)=\alpha_{0} f_{1}(x)$. Such an $x^{\prime}$ will exist if $\alpha_{0} f_{1}(x)$ lies in $\operatorname{Im} f_{1}^{\prime}=\operatorname{Ker} f_{0}^{\prime}$, which it does since $f_{0}^{\prime} \alpha_{0} f_{1}=\alpha f_{0} f_{1}=0$. The same procedure defines all the subsequent $\alpha_{i}$ ‘s.
If we have another chain map extending $\alpha$ given by maps $\alpha_{i}^{\prime}: F_{i} \rightarrow F_{i}^{\prime}$, then the differences $\beta_{i}=\alpha_{i}-\alpha_{i}^{\prime}$ define a chain map extending the zero map $\beta: H \rightarrow H^{\prime}$. It will suffice to construct maps $\lambda_{i}: F_{i} \rightarrow F_{i+1}^{\prime}$ defining a chain homotopy from $\beta_{i}$ to 0 , that is, with $\beta_{i}=f_{i+1}^{\prime} \lambda_{i}+\lambda_{i-1} f_{i}^{\prime}$. The $\lambda_{i}$ ‘s are constructed inductively by a procedure much like the construction of the $\alpha_{i}^{\prime}$ ‘s. When $i=0$ we let $\lambda_{-1}: H \rightarrow F_{0}^{\prime}$ be zero, and then the desired relation becomes $\beta_{0}=f_{1}^{\prime} \lambda_{0}$. We can achieve this by letting $\lambda_{0}$ send a basis element $x$ to an element $x^{\prime} \in F_{1}^{\prime}$ such that $f_{1}^{\prime}\left(x^{\prime}\right)=\beta_{0}(x)$. Such an $x^{\prime}$ exists since $\operatorname{Im} f_{1}^{\prime}=\operatorname{Ker} f_{0}^{\prime}$ and $f_{0}^{\prime} \beta_{0}(x)=\beta f_{0}(x)=0$. For the inductive step we wish to define $\lambda_{i}$ to take a basis element $x \in F_{i}$ to an element $x^{\prime} \in F_{i+1}^{\prime}$ such that $f_{i+1}^{\prime}\left(x^{\prime}\right)=\beta_{i}(x)-\lambda_{i-1} f_{i}(x)$. This will be possible if $\beta_{i}(x)-\lambda_{i-1} f_{i}(x)$ lies in $\operatorname{Im} f_{i+1}^{\prime}=\operatorname{Ker} f_{i}^{\prime}$, which will hold if $f_{i}^{\prime}\left(\beta_{i}-\lambda_{i-1} f_{i}\right)=0$. Using the relation $f_{i}^{\prime} \beta_{i}=\beta_{i-1} f_{i}$ and the relation $\beta_{i-1}=f_{i}^{\prime} \lambda_{i-1}+\lambda_{i-2} f_{i-1}$ which holds by induction, we have
$$
\begin{aligned}
f_{i}^{\prime}\left(\beta_{i}-\lambda_{i-1} f_{i}\right) &=f_{i}^{\prime} \beta_{i}-f_{i}^{\prime} \lambda_{i-1} f_{i} \
&=\beta_{i-1} f_{i}-f_{i}^{\prime} \lambda_{i-1} f_{i}=\left(\beta_{i-1}-f_{i}^{\prime} \lambda_{i-1}\right) f_{i}=\lambda_{i-2} f_{i-1} f_{i}=0
\end{aligned}
$$
EMATH41120-WE01/MATH4161-WE01 COURSE NOTES ：
Now we return to topology. Given a space $X$ and an abelian group $G$, we define the group $C^{n}(X ; G)$ of singular $n$-cochains with coefficients in $G$ to be the dual group $\operatorname{Hom}\left(C_{n}(X), G\right)$ of the singular chain group $C_{n}(X)$. Thus an $n$-cochain $\varphi \in C^{n}(X ; G)$ assigns to each singular $n$-simplex $\sigma: \Delta^{n} \rightarrow X$ a value $\varphi(\sigma) \in G$. Since the singular $n$-simplices form a basis for $C_{n}(X)$, these values can be chosen arbitrarily, hence $n$-cochains are exactly equivalent to functions from singular $n$-simplices to $G$.
The coboundary map $\delta: C^{n}(X ; G) \rightarrow C^{n+1}(X ; G)$ is the dual $\partial^{*}$, so for a cochain $\varphi \in C^{n}(X ; G)$, its coboundary $\delta \varphi$ is the composition $C_{n+1}(X) \stackrel{\partial}{\longrightarrow} C_{n}(X) \stackrel{\varphi}{\longrightarrow} G$. This means that for a singular $(n+1)$-simplex $\sigma: \Delta^{n+1} \rightarrow X$ we have
$$
\delta \Phi(\sigma)=\sum_{i}(-1)^{i} \Phi\left(\sigma \mid\left[v_{0}, \cdots, \hat{v}{i}, \cdots, v{n+1}\right]\right)
$$