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Math8306课程简介
Instructor: Anar Akhmedov
Lectures: MWF 11.15am – 12.05pm in Vincent Hall 213.
E-mail: [email protected]
Office Hours: Monday 12.40 – 2.20pm. My office is in room 355 of the Vincent Hall.
Prerequisites: Math 8301 or instructor’s consent.
Textbook: Algebraic Topology, by Allen Hatcher. The textbook is available at the University bookstore, and also on reserve in the Mathematics Library. Our textbook is also available free online, at http://www.math.cornell.edu/~hatcher/AT/ATpage.html
Prerequisites
Course Outline: This is a first course in algebraic topology. The fall semester we plan to cover Chapters 0 – 3 of the textbook. The main topics are: homotopy, fundamental group, covering spaces, homology, and cohomology.
Web page: http://www.math.umn.edu/~akhmedov/MATH8306.html.
Grading: The course grade will be based on seven homework assignments, in class midterm, and a comprehensive take-home final, with the following weights:
42% Homework25% Midterm (in class)33% Take Home Final
Exams: There will be a comprehensive take-home final examination (which will worth 33% of the final course grade) and in class midterm on October 26 (which will worth 25% of the final course grade). TAKE HOME FINAL
DUE: by noon Monday, December 21
Homework: There will be 7 homeworks in this course, each worth 6 points. Homework will be a fundamental part of this course, and will be worth 42 points (42% of the course grade). NO LATE HOMEWORK WILL BE ACCEPTED. The first homework assignment will be due on September 23. Please staple your homework before handing it in. If you have questions about the homework, it is best to ask during my office hours.
Math8306 Algebraic topology HELP（EXAM HELP， ONLINE TUTOR）
Numbered exercises are from Hatcher’s “Algebraic Topology.”
In class, we defined subdivision maps $s_n^i: \Delta[n+1] \rightarrow \Delta[n] \times[0,1]$ for $0 \leq i \leq n$ by
$$
s_n^i\left(t_1, \cdots, t_{n+1}\right)=\left(\left(t_1, \ldots, \widehat{t_{i+1}}, \ldots, t_{n+1}\right), t_{i+1}\right) .
$$
Show that these satisfy the relations
$s_n^i d_{n+1}^j= \begin{cases}\left(d_n^{j-1}, i d\right) \circ s_{n-1}^i & \text { if } ij\end{cases}$
$s_n^0 d_{n+1}^0=i_0$
$s_n^n d_{n+1}^0=i_1$
$s_n^{i-1} d_{n+1}^i=s_n^i d_{n+1}^i$ for $i \geq 1$.
Use this to show that the operator $h: C_n(X) \rightarrow C_{n+1}(X \times[0,1])$ given by
$$
h\left(\sum a_\sigma \sigma\right)=\sum a_\sigma \sum_{i=0}^n(-1)^i(\sigma, i d) \circ s_n^i
$$
satisfies $\partial h(x)+h \partial \sigma(x)=i_0(x)-i_1(x)$.
To show that the subdivision maps satisfy the given relations, we will use the following commutative diagram for the face maps and degeneracy maps of simplices:
where $\$ 0 \backslash$ leq i\leq $n-1 \$, \$ 0 \backslash$ leq $j \backslash$ leq $n \$$, and $\$ i \backslash e q j-1 \$$.
Now we can prove each of the given relations:
For $\$ i<j-1 \$$, we have
$$
\begin{aligned}
s_n^i d_{n+1}^j & =\left(d_n^{j-1}, i d\right) \circ s_{n-1}^i \circ d_{n+1}^j \
& =\left(d_n^{j-1}, i d\right) \circ d_n^i \circ s_n^{j-1} \
& =d_{n-1}^{i-1} \circ\left(d_n^{j-1}, i d\right) \circ s_{n-1}^i \
& =\left(d_n^{j-1}, i d\right) \circ s_{n-1}^i \circ d_n^{i-1} \
& =\left(d_n^{j-1}, i d\right) \circ s_{n-1}^i \circ d_{n+1}^{j-1} .
\end{aligned}
$$
For $\$ i>j \$$, we have
$$
\begin{aligned}
s_n^i d_{n+1}^j & =\left(d_n^j, i d\right) \circ s_{n-1}^{i-1} \circ d_{n+1}^j \
& =\left(d_n^j, i d\right) \circ d_n^{i-1} \circ s_n^j \
& =d_{n-1}^i \circ\left(d_n^j, i d\right) \circ s_{n-1}^{i-1} \
& =\left(d_n^j, i d\right) \circ s_{n-1}^{i-1} \circ d_n^i \
& =\left(d_n^j, i d\right) \circ s_{n-1}^{i-1} \circ d_{n+1}^i
\end{aligned}
$$
For $\$ \mathrm{i}=0 \$$, we have
$$
\begin{aligned}
s_n^0 d_{n+1}^0 & =\left(d_n^0, i d\right) \circ s_{n-1}^0 \circ d_{n+1}^0 \
& =\left(d_n^0, i d\right) \circ i_0 \
& =i_0
\end{aligned}
$$
For $\$ i=n \$$, we have
$$
\begin{aligned}
s_n^n d_{n+1}^0 & =\left(d_n^0, i d\right) \circ s_{n-1}^n \circ d_{n+1}^0 \
& =\left(d_n^0, i d\right) \circ i_1 \
& =i_1
\end{aligned}
$$
Let $C_*$ be the chain complex with $$ C_n= \begin{cases}\mathbb{Z} & \text { if } n=1, \\ 0 & \text { otherwise }\end{cases} $$ Let $D_*$ be the chain complex with $$ D_n= \begin{cases}\mathbb{Z} & \text { if } n=0,1, \\ 0 & \text { otherwise },\end{cases} $$ such that the boundary map $\partial: D_1 \rightarrow D_0$ sends $m$ to $2 m$. Show that the natural projection $\pi: D_* \rightarrow C_*$ is a map of chain complexes and it induces the zero map $H_*\left(D_*\right) \rightarrow H_*\left(C_*\right)$. Show that there is no chain homotopy $h$ with $\partial h+h \partial=\pi$ (from $\pi$ to zero).
We first show that $\pi$ is a map of chain complexes. Let $\sigma=[a_0,a_1]$ be a generator of $C_1$, then $\pi(\sigma)=[a_0] \in D_0$ and $\partial(\pi(\sigma))=0$. Hence, $\pi(\partial(\sigma))=\partial(\pi(\sigma))=0$, and $\pi$ is indeed a map of chain complexes.
Next, we compute the homology of $C_$ and $D_$. Since $C_n=0$ for $n\neq 1$, we have $H_n(C_*)=\begin{cases} \mathbb{Z} &\text{if } n=1, \ 0 &\text{otherwise.}\end{cases}$
Similarly, since $D_n=0$ for $n\notin {0,1}$, we have $H_n(D_*)=\begin{cases} \mathbb{Z} &\text{if } n=0,1, \ 0 &\text{otherwise.}\end{cases}$
The map $\pi$ is clearly surjective on $D_1$, and hence induces the zero map on homology $H_1(D_)\to H_1(C_)$. Moreover, $\ker\pi=\mathbb{Z}$ is generated by $2$, and hence the map $\pi$ induces the zero map on $H_0$ as well.
Finally, we show that there is no chain homotopy $h$ with $\partial h+h \partial=\pi$ (from $\pi$ to zero). Suppose such an $h$ exists. Since $h$ is a chain homotopy, it must satisfy $h_1(D_1)=0$ and $h_0(D_0)=0$. Since $\pi$ is surjective on $D_1$, there exists $m\in D_1$ such that $\pi(m)=\sigma=[1,1]\in C_1$. Then we have \begin{align*} \sigma&=\pi(m)\ &=(\partial h+h\partial)(m)\ &=\partial(h(m))+h(\partial m)\ &=h_0(2m)+h_1(m)\ &=2h_0(m)+h_1(m). \end{align*} Since $h_0(m)$ and $h_1(m)$ are integers, it follows that $\sigma$ is an even integer, which is a contradiction since $[\sigma]=1$ in $H_1(C_*)$. Therefore, there is no chain homotopy $h$ with $\partial h+h \partial=\pi$.
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