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Let $X$ be a path-connected topological space and $b, b^{\prime} \in X$. We write $I=[0,1]$. (a) Given a path $w: I \rightarrow X$ with $w(0)=b$ and $w(1)=b^{\prime}$, describe the map $$ w_{\sharp}: \pi_1(X, b) \rightarrow \pi_1\left(X, b^{\prime}\right) $$ induced by $w$, and prove that it is well-defined and an isomorphism. How does this isomorphism depend on $w$ ? [You may assume that, if $u$ and $v$ are paths in $X$ such that $u(1)=v(0)$ and $u^{\prime}$ is homotopic to $u$ and $v^{\prime}$ is homotopic to $v$ relative to $\partial I$, then uv and $u^{\prime} v^{\prime}$ are homotopic relative to $\partial I$.] (b) Given loops $\ell$ and $\ell^{\prime}$ in $X$ based at $b$, we say that $\ell$ and $\ell^{\prime}$ are freely homotopic if they are homotopic when viewed as maps from $S^1$ to $X$, not fixing $b$. Give an algebraic necessary and sufficient condition for $\ell$ and $\ell^{\prime}$ to be freely homotopic in terms of their homotopy classes in $\pi_1(X, b)$. (c) Give an example of a space $X$ and distinct elements $l, m \in \pi_1(X, b)$ such that $l$ and $m$ are freely homotopic.

Find all integer solutions to the equation $y^2+74=x^3$. (You may assume that $h_{𝐐(\sqrt{-74})}=10$.) Solution: Let $K=𝐐(\sqrt{-74})$. It turns out that $h_K=10$. In particular, $\mathcal{O}_K$ does not have unique factorisation. The equation factors in $\mathcal{O}_K$ as $(y+\sqrt{-74})(y-\sqrt{-74})=x^3$. We do not have unique factorisation into elements of $\mathcal{O}_K$, only into ideals, so we think of this as an equation \[\tag{1} (y+\sqrt{-74})(y-\sqrt{-74})=(x)^3 \] of ideals. We are going to prove that the two ideals on the left are coprime. Suppose some prime ideal $\mathfrak{p}$ divides both terms on the LHS. Then $y+\sqrt{-74}, y-\sqrt{-74} \in \mathfrak{p}$, and so, taking the difference, $2 \sqrt{-74} \in \mathfrak{p}$. Therefore $\mathfrak{p} \mid(2 \sqrt{-74})$. (Here, of course, we are using the fact that containment and division of ideals are the same thing, Theorem 5.2.) Taking norms, we have \[\tag{12.2} N(\mathfrak{p}) \mid N(2 \sqrt{-74})=2^3 \cdot 37 . \] Also, since $\mathfrak{p} \mid(y+\sqrt{-74})$, we have $\mathfrak{p} \mid(x)^3$ and so \[\tag{12.3} N(\mathfrak{p}) \mid N\left((x)^3\right)=x^6 . \] We claim that neither 2 nor 37 divides $x$. If $2 \mid x$ then $8 \mid x^3$, so $y^2=x^3-74 \equiv 2\pmod 4$, a contradiction. If $37 \mid x$ then $37 \mid y$, and so $37^2 \mid y^2-x^3=74$. This is also a contradiction. From these facts and $(12.2),(12.3)$ we have $N(\mathfrak{p})=1$, which is impossible; therefore we are forced to conclude that $\mathfrak{p}$ does not exist, so the ideals $(y+\sqrt{-74}),(y-\sqrt{-74})$ are indeed coprime. Now we return to (12.1). By unique factorisation of ideals, both $(y+\sqrt{-74})$ and $(y-\sqrt{-74})$ are cubes of ideals. Suppose that $(y+\sqrt{-74})=\mathfrak{a}^3$. In particular, $[\mathfrak{a}]^3$ is trivial in the class group. However, we know that $h_K=10$, that is to say the class group has order 10. Therefore $[\mathfrak{a}]=[\mathfrak{a}]^{10}$ is trivial, or in other words $\mathfrak{a}$ is a principal ideal. Thus we have an equation \[ (y+\sqrt{-74})=(a+b \sqrt{-74})^3 \] for some $a, b \in \mathbf{Z}$. This means that \[ y+\sqrt{-74}=u(a+b \sqrt{-74})^3 \] in $\mathcal{O}_K$, where $u$ is a unit. The only units are $\pm 1$; by replacing $a, b$ with $-a,-b$ if necessary, we may in fact assume that $u=1$. Expanding out and comparing coefficients of $\sqrt{-74}$ (which, of course, is irrational) we obtain \[ y=a\left(a^2-3×74b^2\right), \quad b\left(3 a^2-74b^2\right)=1 . \] The second of these implies that $b= \pm 1$ and hence that $3 a^2-74= \pm 1$, so $a=±5,y=±197×5=±985,x=99$.

Theorem 13.21. Let $G$ be a free group on a set $X$. Then $J$ is a free $ℤ G$-module with basis $\{x-1: x ∈ X\}$. Proof. Induct on the length of word. See Weibel Page 169, Proposition 6.2.6. Corollary 13.22. If $G$ is a free group on $X$, then $ℤ$ has free resolution \[ 0 → J → ℤ G → ℤ → 0 \] Therefore, $H_n(G, A)=H^n(G, A)=0$ for $n ≠ 0,1$, and $H_0(G, ℤ) ≅ H^0(G, ℤ) ≅ ℤ$, while \[ H_1(G, ℤ) ≅ \bigoplus_{x ∈ X} ℤ,   H^1(G, ℤ) ≅ \prod_{x ∈ X} ℤ \] Proof. $H_*(G ; A)$ is the homology of $0 →J ⊗_{ℤ G} A → A → 0$, and $H^*(G ; A)$ is the cohomology of $0 → A → \operatorname{Hom}_G(J, A) → 0$. For $A=ℤ$, the differentials are zero.

an orthonormal basis for Bergman space $f_n(z) = \sqrt{\frac{n+1}{\pi}} z^n$ To prove $f_n$ is an orthonormal sequence: $dxdy=rdrdθ$ $\int_{|z|≤1}|f_n|^2dxdy=\int_0^{2\pi}\int_0^1\frac{n+1}{\pi} r^{2n}rdrdθ=1$ For $n≠m$, $\int_{|z|≤1}f_n\overline{f_m}dxdy=\sqrt{\frac{n+1}{\pi}\frac{m+1}{\pi}}\int_0^{2\pi}\int_0^1(re^{iθ})^n(re^{-iθ})^mrdrdθ\\=\sqrt{\frac{n+1}{\pi}\frac{m+1}{\pi}}(\int_0^{2\pi}e^{(n-m)iθ}dθ)(\int_0^1r^{n+m+1}dr)=0$ Proof it is spanning: ??

Two possibilities can occur, i.e either $f$ is bounded or unbounded. If it is bounded then $f$ is continuous and hence $\ker f=f^{-1}(0)$ which is closed. Suppose $f$ is unbounded. Then there exists a sequence $\{x_n\}$ with $x_n→0$ and $f(x_n)=1$ for all $n$. Now choose any $y∈ X$. Then $y-f(y)x_n$ is in $\ker f$, since $f(y-f(y)x_n)=f(y)-f(y)=0.$ $$\lim_{n→∞}x_n=0\implies y=\lim_{n→∞}(y-f(y)x_n)$$where the elements in the sequence $\{y-f(y)x_n\}$ are in $\ker f$. So $\ker f$ is dense in $X$.

Part II Lecture Notes: Algebraic Topology, James Lingard Let $X$ be a simplicial complex which is the union $A \cup B$ of two subcomplexes. Then $$ \cdots \longrightarrow H_{i}(A \cap B) \longrightarrow H_{i}(A) \oplus H_{i}(B) \longrightarrow H_{i}(X) \longrightarrow H_{i-1}(A \cap B) \longrightarrow \cdots $$ is a long exact sequence. What are the three homomorphisms? • The homomorphism $H_{i}(A \cap B) \rightarrow H_{i}(A) \oplus H_{i}(B)$ is the obvious pair of inclusions. • The homomorphism $H_{i}(A) \oplus H_{i}(B) \rightarrow H_{i}(X)$ is given by $(x, y) \mapsto x-y$. • The homomorphism $H_{i}(X) \rightarrow H_{i-1}(A \cap B)$ is the "boundary map" constructed as follows. \[\begin{CD} 0@>>>C_i(A\cap B)@>>>C_i(A)\oplus C_i(B)@>>>C_i(X)@>>>0\\ &@VV\partial V@VV\partial V@VV\partial V\\ 0@>>>C_{i-1}(A\cap B)@>>>C_{i-1}(A)\oplus C_i(B)@>>>C_{i-1}(X)@>>>0\\ \end{CD}\] For any element of $H_{i}(X)$, take a representative cycle $z \in Z_{i}(X)$ and choose an element $d \in C_{i}(A) \oplus C_{i}(B)$ which maps to $z$. Then the equivalence class of $z$ maps to the equivalence class of the (unique) inverse image $c \in Z_{i-1}(A \cap B)$ for $\partial(d)$.

$\DeclareMathOperator{\sgn}{sgn}\DeclareMathOperator{\sinc}{sinc}\DeclareMathOperator{\fp}{fp}\DeclareMathOperator{\pv}{pv}$For each of the following functions from $ℝ$ into $ℂ$ calculate the Fourier transform: (i) $\cos , \sin , \cos^2, \cos^k$ for $k ∈ ℕ$. Solution. Use $\hat1=2πδ_0$. By translation rule $\widehat{e_{𝗂h}}=τ_{-h}(2πδ_0)=2πδ_h$ \[ \widehat{𝖾^{𝗂x}}=2πδ_1\\ \widehat{𝖾^{-𝗂x}}=2πδ_{-1} \] so \begin{align*} \widehat{\cos}&=πδ_1+πδ_{-1}\\ \widehat{\sin}&=\fracπ𝗂(δ_1-δ_{-1})=𝗂π(δ_{-1}-δ_1)\\ \cos^2(x)&=\frac12+\frac14(𝖾^{2𝗂x}+𝖾^{-2𝗂x})\\ \widehat{\cos^2}&=πδ_0+\fracπ2(δ_2+δ_{-2})\\ \cos^k(x)&=\frac1{2^k}\sum_{n=0}^{k}\binom{k}{n}𝖾^{𝗂nx}𝖾^{-𝗂(k-n)x}\\ \widehat{\cos^k}&=\frac{π}{2^{k-1}}\sum_{n=0}^{k}\binom{k}{n}δ_{k-2n} \end{align*} (ii) $\sinc$ Solution. $\widehat{1_{[-1,1]}}=2\sinc$. By Fourier inversion $\widehat{\sinc}=2π⋅\frac12\widetilde{1_{[-1,1]}}=π1_{[-1,1]}$. (iii) $H$ by calculating first the Fourier transform of $H^ε(x)=𝖾^{-εx}H(x)$ for $ε>0$ ($H$ is Heaviside's function). Solution. \[\int_ℝH^ε(x)𝖾^{-𝗂xξ}𝖽ξ=\int_{0}^{∞}𝖾^{-(𝗂ξ+ε)x}𝖽ξ=\frac1𝗂\lim_{x→∞}\frac{1-𝖾^{-(𝗂ξ+ε)x}}{ξ-𝗂ε}→-𝗂(ξ-𝗂0)^{-1}\] To prove $H^ε→H$ as $ε→0$ in $𝒮'(ℝ)$, \[|T_{H^ε}(ϕ)-T_H(ϕ)|=\left|\int_{-∞}^{∞}(H^ε(x)-H(x))ϕ(x)𝖽x\right|≤\int_{0}^{∞}|𝖾^{-εx}-1||ϕ(x)|𝖽x→0\] so by $𝒮'$-continuity of $ℱ$, $\hat{H}(ξ)=\lim_{ε→0^+}\widehat{H^ε}(ξ)=-𝗂(ξ-𝗂0)^{-1}$. ∎ Deduce that \[ (ξ-𝗂 0)^{-1}=\pv\left(\frac{1}{ξ}\right)+π 𝗂 δ_0, \] where the distribution on the left-hand side again was defined on B4.3 Problem Sheet 4. In both deductions you should carefully explain why the considered distributions are tempered. Solution. By lecture $\hat{H}(ξ)=-𝗂\pv(\frac1{ξ})+π δ_0(ξ)$, so the identity holds. $(ξ-𝗂 0)^{-1}$ is compactly supported, so it is tempered, since $ℰ′(ℝ)⊂𝒮'(ℝ)$. ∎ (iv) $x H(x)=x^+,|x|, \sin |x|$. Solution. By differentiation rule $\widehat{xH}(ξ)=𝗂\hat{H}'(ξ)=-\fp(\frac1{ξ^2})+𝗂π δ'_0(ξ)$ $|x|=x^++(-x)^+\implies\widehat{|x|}(ξ)=-2\fp(\frac1{ξ^2})$ Using $\sin|x|=\sgn(x)\sin(x)$, by translation rule $\widehat{\sin|x|}=\frac{1}{2𝗂}(τ_{-1}\widehat{\sgn}-τ_1\widehat{\sgn})$ Here $\widehat{\sgn}=\hat H-\hat{\tilde{H}}=\hat H-\tilde{\hat{H}}=-2𝗂\pv(\frac1ξ)$ $\implies\widehat{\sin|x|}=-τ_{-1}\pv(\frac1ξ)+τ_1\pv(\frac1ξ)=\pv(\frac1{ξ+1})-\pv(\frac1{ξ-1})$ ∎ Deduce that \[ ℱ_{x → ξ}\left(\fp\left(\frac{1}{x^2}\right)\right)=-π|ξ|, \] where the finite part distribution was defined on B4.3 Sheet 4. Solution. By Fourier inversion, $\widehat{|x|}(ξ)=-2\fp(\frac1{ξ^2})\impliesℱ_{x → ξ}\left(\fp\left(\frac{1}{x^2}\right)\right)=2π\frac1{-2}|-ξ|=-π|ξ|$ $\fp\left(\frac{1}{x^2}\right)$ is compactly supported, so it is tempered, since $ℰ′(ℝ)⊂𝒮'(ℝ)$.

Let $X$ be a normed space, $Y$ a Banach space and let $(T_k) ⊂ ℬ(X, Y)$ be a sequence of compact operators. (a) Let $(x_n)$ be a bounded sequence in $X$. Show that there exists a subsequence $x_{n_j}$ so that $T_k(x_{n_j})$ converges for every $k ∈ ℕ$ (b) Hence or otherwise show that if $T_k$ converges in $ℬ(X, Y)$ to an operator $T$ then $T$ is also a compact operator. Proof. (a) Let $\{x_n\}$ be a bounded sequence in $X$. By compactness, there exists a subsequence of $\{x_n\}$, which we will label $\{x_{n(1, r)}\}$($=\{x_{n(1, r)}\}_{r=1}^{∞}$), such that the sequence $\{T_1 x_{n(1, r)}\}$ converges. Similarly, there exists a subsequence $\{x_{n(2, r)}\}$ of $\{x_{n(1, r)}\}$ such that $\{T_2 x_{n(2, r)}\}$ converges. Also, $\{T_1 x_{n(2, r)}\}$ converges since it is a subsequence of $\{T_1 x_{n(1, r)}\}$. Repeating this process inductively, we see that for each $j ∈ ℕ$ there is a subsequence $\{x_{n(j, r)}\}$ with the property: for any $k ≤ j$ the sequence $\{T_k x_{n(j, r)}\}$ converges. Taking the diagonal sequence $n(r)=n(r, r)$, for $r ∈ ℕ$, we obtain a single subsequence $\{x_{n(r)}\}$ with the property that, for each fixed $k ∈ ℕ$, the sequence $\{T_k x_{n(r)}\}$ converges as $r → ∞$. (b) We will now show that the sequence $\{T x_{n(r)}\}$ converges. We do this by showing that $\{T x_{n(r)}\}$ is a Cauchy sequence, and hence is convergent since $Y$ is a Banach space. Let $ϵ>0$ be given. Since the subsequence $\{x_{n(r)}\}$ is bounded there exists $M>0$ such that $\left\|x_{n(r)}\right\| ≤ M$, for all $r ∈ ℕ$. Also, since $\left\|T_k-T\right\| → 0$, as $k → ∞$, there exists an integer $K ≥ 1$ such that $\left\|T_K-T\right\|<ϵ / 3 M$. Next, since $\{T_K x_{n(r)}\}$ converges there exists an integer $R ≥ 1$ such that if $r, s ≥ R$ then $\left\|T_K x_{n(r)}-T_K x_{n(s)}\right\|<ϵ / 3$. But now we have, for $r, s ≥ R$, \[\left\|T x_{n(r)}-T x_{n(s)}\right\|<\left\|T x_{n(r)}-T_K x_{n(r)}\right\|+\left\|T_K x_{n(r)}-T_K x_{n(s)}\right\| +\left\|T_K x_{n(s)}-T x_{n(s)}\right\|<ϵ,\] which proves that $\{T x_{n(r)}\}$ is a Cauchy sequence.