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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 07, 2017}}}  %
\fancyhead[R]{{\emph{6. 4. 2017}}}                 % Nastavení pro titulní stranu
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\ex (39 in total) Use theorem "$f\colon S^n\to S^n$ s.t. $f(-x)=-f(x) \Rightarrow \operatorname{deg}(f)$ is odd" to prove that for $f\colon S^n \to \R$ there exists $x\in S^n\colon f(x)=f(-x)$.

\begin{proof}
We will use proof by contradiction. Suppose that $f(x)-f(-x)$ is always non-zero and define $g(x)=\dfrac{f(x)-f(-x)}{||f(x)-f(-x)||}$. Then obviously $g(-x)=-g(x)$. Since $g\colon S^n\to S^{n-1}\subset S^n$ is odd , the map $g$ has odd degree, but $g(S^n)\subsetneq S^n$ which implies $\operatorname{deg}(g)=0$ ($g$ is homotopic to a constant map). Now we use well known fact that zero is even number, some say that it is the evennest number of them all (why?) and conclude desired contraction.
\end{proof}

\ex (40 in total) Let $\varphi\in C^k(X;R), \psi\in C^l(Y;R)$. Prove $\delta(\varphi\cup \psi)=\delta\varphi\cup\psi+(-1)^{k}\varphi\cup\delta\psi$. Use $\tau=[e_0,\dots,e_{k+l+1}]\in C_{k+l+1}(X)$.

\begin{proof}

Easily work out 
\begin{align*}
    \delta(\varphi\cup\psi)(\tau)&=(\varphi\cup\psi)(\delta\tau)=(\varphi\cup\psi)\Big(\sum_{i=0}^{k+l+1}(-1)^i \tau/[e_0,\dots,\hat{e_i},\dots,e_{k+l+1}]\Big)=\\
=&\sum_{i=0}^k (-1)^i \varphi(\tau/[e_0,\dots,\hat{e_i},\dots,e_{k+1}])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])+\\
+&\sum_{i=k+1}^{k+l+1} (-1)^i\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_k,\dots,\hat{e_i},\dots,e_{k+l+1}]).
\end{align*}
Now, the right hand side of the formula, the first part gives
\begin{align*}
    (\delta\varphi\cup\psi)(\tau)&=\delta\varphi(\tau/[e_0,\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=\\
                                 &=\varphi(\delta\tau/[e_0,\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=\\
                                 &=\sum_{i=0}^{k+1}(-1)^i \varphi(\delta\tau/[e_0,\dots,\hat{e_i},\dots,e_{k+1}])\cdot \psi(\tau/[e_{k+1},\dots,e_{k+l+1}]).
\end{align*}
The second part is
\begin{align*}
    (-1)^k(\varphi\cup\delta\psi)(\tau)&=(-1)^k\varphi(\tau/[e_0,\dots,e_{k}])\delta \psi(\tau/[e_{k},\dots,e_{k+l+1}])=\\
    &=\sum_{j=0}^{l+1}(-1)^{j+k} \varphi(\delta\tau/[e_0,\dots,e_{k}])\cdot \psi(\tau/[e_{k},\dots,\hat{e}_{k+j},\dots,e_{k+l+1}]).
\end{align*}
Now, the last summand of the first part plus the first summand of the second part yields
\begin{align*}
(-1)^{k+1}&\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])+\\
+(-1)^k&\varphi(\tau/[e_0,\dots,e_k])\psi(\tau/[e_{k+1},\dots,e_{k+l+1}])=0,    
\end{align*}
and we are done, $\mathsf{LHS=RHS}$.

\end{proof}

\ex (41 in total) Prove $a\cup b=\Delta^*(a\times b)$, where $a\in H^k(X),b\in H^l(X),\Delta\colon X\to X\times X,x\mapsto (x,x)$ and $\times$ is cross product defined $\alpha\times \beta=p_X^* \alpha \cup p_Y^*\beta$ ($p_X,p_Y$ are projections from $X\times Y$).

\begin{proof}
By the following diagram
\[
\xymatrix{
X \ar@{->}[rr]^{\Delta_X}  & & X\times X \ar@{->}[ld]_{p_X=p_1} \ar@{->}[d]^{p_Y=p_1} & \\
& X & X}
\]
we have that $p_X\Delta$ is $\id$. So, compute
$$\Delta^*(a\times b)=\Delta^*(p_X^*a\cup p_Y^*b)=\Delta^*(p_X^*a)\cup\Delta^*(p_Y^*b)=(p_X\Delta)^* a\cup (p_Y\Delta)^*b,$$
and we are done. The thing is that cup product is natural.
\end{proof}

\ex (42 in total) Compute the structure of graded algebra $H^*(S^n\times S^n;\Z)$ for $n$ even and $n$ odd. Use the following:\par 
If $H^n(Y;R)$ is free finitely generated group for all $n$ and $(X,A),Y$ are CW-complexes, then
$$\times\colon H^*(X,A;R)\otimes H^*(Y;R)\to H^*(X\times Y,A\times Y;R)$$
is an isomorphism of graded rings.

\begin{proof}
We will omit writing the $\Z$ coefficients.\par 
Now, $H^*(S^n)\otimes H^*(S^n)\to H^*(S^n\times S^n)$ and we know that for spheres $H^0=\Z$ with generator $1$ and $H^n=\Z$, denote generator $a$. Also, $a\cup a\in H^{2n}=0, a\cup a=0$, so we get $\Z[a]/\langle a^2\rangle$ and $\operatorname{deg}(a)=n$. We can write the same for the second, so denote the other generator $b$ and have $\operatorname{deg}(b)=n$ and we have $\Z[b]/\langle b^2\rangle$.\par 
Now we compute tensor product $\Z[a]/\langle a^2\rangle\otimes \Z[b]/\langle b^2\rangle$, we have four generators: $1_a\otimes 1_b,a\otimes 1_b,1_a\otimes b,a\otimes b$, we will denote them $1,c,d,c\cdot d$. Compute
$$(a\otimes 1_b)\cdot (1_a\otimes b)=(-1)^{0\cdot 0}(a\cdot 1_a)\otimes (1_b\cdot b)=a\otimes b,$$
because $0$ is an idempotent element, i.e. $0\cdot 0=0$, and $(-1)^n=1$ for $n$ even, again, as in the first exercise, we use Evenness of Zero. (We refer the reader to "Principia Mathematica" \textit{Whitehead, Russell},(1910,1912,1913).) Continue with computation
$$(1_a\otimes b)\cdot (a\otimes 1_b)=(-1)^{n\cdot n}(1_a\cdot a)\otimes (b\cdot 1_b)=(-1)^na\otimes b,$$
so the algebra we get is $H^*(S^n\times S^n)=\Z[c,d]/\langle c^2,d^2,dc-(-1)^ncd\rangle$. For $n$  even we have $dc=cd$.
\end{proof}

\newpage

\ex (43 in total) Prove that there is no multiplication on even dimensional spheres. Multiplication on the sphere $S^n$ is a map $m\colon S^n\times S^n\to S^n$ such that there is an element $1\in S^n$ satisfying $m(x,1)=x,m(1,x)=x$.\par 
Hint: compute $m^*\colon H^*(S^n)\to H^*{(S^n\times S^n)}$, describe two rings.
\begin{proof}
We have $H^*(S^n)=\Z[\gamma]/\langle\gamma^2\rangle$ and $H^*(S^n\times S^n)=\Z[\alpha,\beta]/\langle\alpha^2,\beta^2\rangle=H^*(S^n)\otimes H^*(S^n)$, because we already know, that $\alpha\beta=\beta\alpha$. Our situation can be described with two diagrams:
\[
\xymatrix{
S^n \ar[r]^{i_1} \ar[dr]_{\id} & S^n\times S^n \ar[d]^m\\
& S^n
}\quad
\xymatrix{
H^{\ast}(S^n) & H^*(S^n\times S^n)  \ar[l]_{i_1^*}\\
& H^*(S^n) \ar[u]_{m^*} \ar[lu]^{\id}
}
\]

Take $m^*(\gamma)=a\alpha+b\beta$ with $a,b\in\Z$ and prove first, that $a=b=1$. Use $m\circ i_1=\id$, so $i_1^*(m^*\gamma)=\gamma$. This gives $i_1^*(a\alpha+b\beta)=\gamma$ and since $i_1^*(a\alpha+b\beta)=a\gamma$, we have $a=1$, same for $b$. Final computation yields
\begin{align*}
0&=m^*(0)=m^*(\gamma^2)=m^*(\gamma\cup\gamma)=m^*\gamma\cup m^*\gamma=\\&=(\alpha+\beta)\cup(\alpha+\beta)=\alpha^2+\alpha\beta+\beta\alpha+\beta^2=0+2\alpha\beta+0\neq 0,
\end{align*}
and that, my friends, is a contradiction.
\end{proof}

\ex (unfinished) (44 in total) Use five lemma to prove that taking any two $\mu$'s iso's, the third $\mu$ is iso as well. (Proving five lemma might be a homework.) Also, show commutativity of the diagram.

\[
\xymatrix{
H^*(X,A) \otimes H^*(Y) \ar[rr] \ar[dd]^{\mu} &  & H^*(X)\otimes H^*(Y)\ar[dd]^{\mu} \ar[ld]\\
 & H^*(A)\otimes H^*(Y)\ar[dd]^<<<<{\mu} \ar[ul]_{\delta^*\otimes\id} & \\
H^*(X\times Y,A\times Y)\ar[rr] &  & H^*(X\times Y)\ar[ld]\\
 & H^*(A\times Y) \ar[lu]_{\delta^*}& 
}
\]

\begin{proof}
We will name the parts of the diagram as follows: back-square, upper-triangle, lower-triangle, left-square, right-square. The triangles come from the long exact sequence of of pairs $(X,A)$ and $(X\times Y, A\times Y)$, the right-square commutativity comes from an inclusion, back-square commutes as well (topology knowledge). The only problematic part is the left-square and it is exercise on computation of connecting homomorphism $\delta^*$.
\end{proof}

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