Fix diagram (and typo)

II_M/algebraic_topology.tex

@@ -128,7 +128,7 @@ \subsection{Cell complexes}
   \]
   where the equivalence relation $\sim$ is the equivalence relation generated by $x\sim f(x)$ for all $x\in S^{n - 1}\subseteq D^n$ (and $\amalg$ is the disjoint union).
 
-  Intuitively, a map $f: S^{n - 1}\to X$ just picks out a subset $X$ that looks like the sphere. So we are just sticking a disk onto $X$ by attaching the boundary of the disk onto a sphere within $X$.
+  Intuitively, a map $f: S^{n - 1}\to X$ just picks out a subset of $X$ that looks like the sphere. So we are just sticking a disk onto $X$ by attaching the boundary of the disk onto a sphere within $X$.
   \begin{center}
     \begin{tikzpicture}
       \draw ellipse (0.5 and 1);
@@ -701,7 +701,7 @@ \subsection{The fundamental group}
   Immediate from our previous lemmas.
 \end{proof}
 
-Often, in mathematics, after defining a term, we give lots of examples of it. Unfortunately, it is rather difficult to prove that a space has a non-trivial fundamental group, until we have developed some relevant machinery. Hence we will have to wait for a while before we have some concrete examples. Instead, we will look at some properties of the fundamental group first.
+Often in mathematics, after defining a term, we give lots of examples of it. Unfortunately, it is rather difficult to prove that a space has a non-trivial fundamental group, until we have developed some relevant machinery. Hence we will have to wait for a while before we have some concrete examples. Instead, we will look at some properties of the fundamental group first.
 
 \begin{defi}[Based space]
   A \emph{based space} is a pair $(X, x_0)$ of a space $X$ and a point $x_0\in X$, the \emph{basepoint}. A \emph{map of based spaces}
@@ -1399,9 +1399,9 @@ \subsection{Covering space}
     \item The stabilizer of $\tilde{x}_0 \in p^{-1}(x_0)$ is $p_*(\pi_1(\tilde{X}, \tilde{x}_0)) \subseteq \pi_1(X, x_0)$.
     \item If $\tilde{X}$ is path connected, then there is a bijection
       \[
-        p_* \pi_1(\tilde{X}, \tilde{x}_0)\backslash \pi_1(X, x_0) \to p^{-1}(x_0).
+        p_* (\pi_1(\tilde{X}, \tilde{x}_0))\backslash \pi_1(X, x_0) \to p^{-1}(x_0).
       \]
-      Note that $p_* \pi_1(\tilde{X}, \tilde{x}_0)\backslash \pi_1(X, x_0)$ is not a quotient, but simply the set of cosets. We write it the ``wrong way round'' because we have right cosets instead of left cosets.
+      Note that $p_*(\pi_1(\tilde{X}, \tilde{x}_0))\backslash \pi_1(X, x_0)$ is not a quotient, but simply the set of cosets. We write it the ``wrong way round'' because we have right cosets instead of left cosets.
   \end{enumerate}
 \end{lemma}
 Note that this is great! If we can find a covering space $p$ and a point $x_0$ such that $p^{-1}(x_0)$ is non-trivial, then we immediately know that $\pi_1(X, x_0)$ is non-trivial!
@@ -1716,7 +1716,7 @@ \subsection{The Galois correspondence}
 
 Now we prove that every subgroup of $\pi_1$ comes from exactly one covering space. What this statement properly means is made precise in the following proposition:
 \begin{prop}
-  Let $(X, x_0)$, $(\tilde{X}_1, \tilde{x}_1)$, $(\tilde{X}_2, \tilde{X}_2)$ be path-connected based spaced, and $p_i: (\tilde{X}_i, \tilde{x}_i) \to (X, x_0)$ be covering maps. Then we have
+  Let $(X, x_0)$, $(\tilde{X}_1, \tilde{x}_1)$, $(\tilde{X}_2, \tilde{x}_2)$ be path-connected based spaced, and $p_i: (\tilde{X}_i, \tilde{x}_i) \to (X, x_0)$ be covering maps. Then we have
   \[
     p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1) = p_{2*} \pi_1(\tilde{X}_2, \tilde{x_2})
   \]
@@ -2209,7 +2209,7 @@ \subsection{Seifert-van Kampen theorem}
   \[
     \pi_1(S^n) \cong \pi_1(\R^n) \underset{\pi_1(S^{n - 1})}{*} \pi_1(\R^n) \cong 1 \underset{\pi_1(S^{n - 1})}{*} 1
   \]
-  It is easy to see this is the trivial group. We can see this directly form the universal property of the amalgamated free product, or note that it is the quotient of $1 * 1$, which is $1$.
+It is easy to see this is the trivial group. We can see this directly from the universal property of the amalgamated free product, or note that it is the quotient of $1 * 1$, which is $1$.
 
   So for $n \geq 2$, $\pi_1(S^n) \cong 1$.
 \end{eg}
@@ -2428,7 +2428,7 @@ \subsection{The effect on \texorpdfstring{$\pi_1$}{pi1} of attaching cells}
 \end{proof}
 The more interesting case is when we have smaller dimensions.
 \begin{thm}
-  If $n = 2$, then the natural map $\pi_1(X, X_0) \to \pi_1(X\cup_f D^n, x_0)$ is \emph{surjective}, and the kernel is $\bra\bra [f] \ket \ket$. Note that this statement makes sense, since $S^{n - 1}$ is a circle, and $f: S^{n - 1} \to X$ is a loop in $X$.
+  If $n = 2$, then the natural map $\pi_1(X, x_0) \to \pi_1(X\cup_f D^n, x_0)$ is \emph{surjective}, and the kernel is $\bra\bra [f] \ket \ket$. Note that this statement makes sense, since $S^{n - 1}$ is a circle, and $f: S^{n - 1} \to X$ is a loop in $X$.
 \end{thm}
 This is what we would expect, since if we attach a disk onto the loop given by $f$, this loop just dies.
 
@@ -2856,7 +2856,7 @@ \subsection{Simplicial complexes}
 
   The boundary of $\sigma$ is usually denoted by $\partial \sigma$, while the interior is denoted by $\mathring{\sigma}$, and we write $\tau \leq \sigma$ when $\tau$ is a face of $\sigma$.
 \end{defi}
-In particular, the interior of a vertex is the vertex itself. Note that this notions of interior and boundary are distinct from the topological notions of interior and boundary.
+In particular, the interior of a vertex is the vertex itself. Note that these notions of interior and boundary are distinct from the topological notions of interior and boundary.
 
 \begin{eg}
   The \emph{standard $n$-simplex} is spanned by the basis vectors $\{\mathbf{e}_0, \cdots, \mathbf{e}_n\}$ in $\R^{n + 1}$. For example, when $n = 2$, we get the following:
@@ -3177,7 +3177,7 @@ \subsection{Simplicial approximation}
 \end{defi}
 
 \begin{prop}
-  $|K| = |K|'$ and $K'$ really is a simplicial complex.
+  $|K| = |K'|$ and $K'$ really is a simplicial complex.
 \end{prop}
 
 \begin{proof}
@@ -3186,12 +3186,12 @@ \subsection{Simplicial approximation}
 
 We now have a slight problem. Even though $|K'|$ and $|K|$ are equal, the identity map from $|K'|$ to $|K|$ is not a simplicial map.
 
-To solve this problem, we can choose any function $K \to V_K$ by $\sigma \mapsto v_\sigma$ with $v_\sigma \in \sigma$, i.e.\ a function that sends any simplex to any of its vertices. Then we can define $g: K' \to K$ by sending $\hat{\sigma} \mapsto v_\sigma$. Then this is a simplicial map, and indeed a simplicial approximation to the identity map $|K'| \to |K|$.
+To solve this problem, we can choose any function $K \to V_K$ by $\sigma \mapsto v_\sigma$ with $v_\sigma \in \sigma$, i.e.\ a function that sends any simplex to any of its vertices. Then we can define $g: K' \to K$ by sending $\hat{\sigma} \mapsto v_\sigma$. Then this is a simplicial map, and indeed a simplicial approximation to the identity map $|K'| \to |K|$. We will revisit this idea later when we discuss homotopy invariance.
 
 The key theorem is that as long as we are willing to perform barycentric subdivisions, then we can always find a simplicial approximation.
 
 \begin{thm}[Simplicial approximation theorem]
-  Le $K$ and $L$ be simplicial complexes, and $f: |K| \to |L|$ a continuous map. Then there exists an $r$ and a simplicial map $g: K^{(r)} \to L$ such that $g$ is a simplicial approximation of $f$. Furthermore, if $f$ is already simplicial on $M\subseteq K$, then we can choose $g$ such that $|g||_M = f|_M$.
+  Let $K$ and $L$ be simplicial complexes, and $f: |K| \to |L|$ a continuous map. Then there exists an $r$ and a simplicial map $g: K^{(r)} \to L$ such that $g$ is a simplicial approximation of $f$. Furthermore, if $f$ is already simplicial on $M\subseteq K$, then we can choose $g$ such that $|g||_M = f|_M$.
 \end{thm}
 
 The first thing we have to figure out is how far we are going to subdivide. To do this, we want to quantify how ``fine'' our subdivisions are.
@@ -3204,9 +3204,9 @@ \subsection{Simplicial approximation}
 
 We have the following lemma that tells us how large our mesh is:
 \begin{lemma}
-  Let $\dim K \leq n$, then
+  Let $\dim K = n$, then
   \[
-    \mu(K^{(r)}) = \left(\frac{n}{n + 1}\right)^r \mu(K).
+    \mu(K^{(r)}) \leq \left(\frac{n}{n + 1}\right)^r \mu(K).
   \]
 \end{lemma} % insert proof
 The key point is that as $r \to \infty$, the mesh goes to zero. So indeed we can make our barycentric subdivisions finer and finer. The proof is purely technical and omitted.
@@ -3647,7 +3647,7 @@ \subsection{Some homological algebra}
     & & D_n & D_{n + 1} \ar [l, "d_{n + 1}"]
   \end{tikzcd}
 \]
-The intuition behind this definition is as follows: suppose $C_{\Cdot} = C_{\Cdot} (K)$ and $D_{\Cdot} = C_{\Cdot}(L)$ for $K, L$ simplicial complexes, and $f_{\Cdot}$ and $g_{\Cdot}$ are ``induced'' by simplicial maps $f, g: K \to L$. How can we detect if $f$ and $g$ are homotopic via the homotopy groups?
+The intuition behind this definition is as follows: suppose $C_{\Cdot} = C_{\Cdot} (K)$ and $D_{\Cdot} = C_{\Cdot}(L)$ for $K, L$ simplicial complexes, and $f_{\Cdot}$ and $g_{\Cdot}$ are ``induced'' by simplicial maps $f, g: K \to L$, where by ``induced'' we mean where $n$-simplices are mapped to $k$-simplices for $k<n$, $f_{\Cdot}=0$. How can we detect if $f$ and $g$ are homotopic via the homotopy groups?
 
 Suppose $H: |K| \times I \to |L|$ is a homotopy from $f$ to $g$. We further suppose that $H$ actually comes from a simplicial map $K \times I \to L$ (we'll skim over the technical issue of how we can make $K \times I$ a simplicial complex. Instead, you are supposed to figure this out yourself in example sheet 3).
 
@@ -3717,7 +3717,7 @@ \subsection{Some homological algebra}
     \item If $a_{\Cdot}: A_{\Cdot} \to C_{\Cdot}$ is a chain map and $f_{\Cdot} \simeq g_{\Cdot}$, then $f_{\Cdot} \circ a_{\Cdot} \simeq g_{\Cdot} \circ a_{\Cdot}$.
     \item If $f: C_{\Cdot} \to D_{\Cdot}$ and $g: D_{\Cdot} \to A_{\Cdot}$ are chain maps, then
       \[
-        g_* \circ f_* = (f_{\Cdot} \circ g_{\Cdot})_*.
+        g_* \circ f_* = (g_{\Cdot} \circ f_{\Cdot})_*.
       \]
     \item $(\id_{C_{\Cdot}})_* = \id_{H_*(C)}$.
   \end{enumerate}
@@ -4293,6 +4293,10 @@ \subsection{Continuous maps and homotopy invariance}
   \end{enumerate}
 \end{prop}
 
+\begin{proof}
+  Omitted.
+\end{proof}
+
 \begin{cor}
   If $f: |K| \to |L|$ is a homeomorphism, then $f_*: H_n(K) \to H_n(L)$ is an isomorphism for all $n$.
 \end{cor}
@@ -4314,7 +4318,7 @@ \subsection{Continuous maps and homotopy invariance}
 \begin{lemma}
   Let $L$ be a simplicial complex (with $|L| \subseteq \R^n$). Then there is an $\varepsilon = \varepsilon(L) > 0$ such that if $f, g: |K| \to |L|$ satisfy $\|f(x) - g(x)\| < \varepsilon$, then $f_* = g_*: H_n(K) \to H_n(L)$ for all $n$.
 \end{lemma}
-The idea of the proof is that if $\|f(x) - g(x)\|$ is small enough, we can barycentrically subdivide $L$ such that we get a simplicial approximation to both $f$ and $g$.
+The idea of the proof is that if $\|f(x) - g(x)\|$ is small enough, we can barycentrically subdivide $K$ such that we get a simplicial approximation to both $f$ and $g$.
 \begin{proof}
   By the Lebesgue number lemma, there is an $\varepsilon > 0$ such that each ball of radius $2\varepsilon$ in $|L|$ lies in some star $\St_L(w)$.
 
@@ -4326,9 +4330,9 @@ \subsection{Continuous maps and homotopy invariance}
   \[
     g(B_\delta(x)) \subseteq B_{2\varepsilon}(y) \subseteq \St_L(w).
   \]
-  Now subdivide $r$ times so that $\mu(K^{(r)}) < \frac{1}{2} \delta$. So for all $v \in V_K(R)$, we know
+  Now subdivide $r$ times so that $\mu(K^{(r)}) < \frac{1}{2} \delta$. So for all $v \in V_{K^(r)}$, we know
   \[
-    \St_{K(r)} (v) \subseteq B_\delta(V).
+    \St_{K^{(r)}} (v) \subseteq B_\delta(v).
   \]
   This gets mapped by \emph{both} $f$ and $g$ to $\St_L(w)$ for the same $w \in V_L$. We define $s: V_{K^{(r)}} \to V_L$ sending $v \mapsto w$.
 \end{proof}
@@ -4443,7 +4447,7 @@ \subsection{Homology of spheres and applications}
       \node at (-1.6, -1.2) [anchor = north east] {$g(x)$};
     \end{tikzpicture}
   \end{center}
-  So we now show no such continuous retraction can exist. Suppose $r: D^n \to \partial D^n$ is a retraction, i.e.\ $r \circ i \simeq \id: \partial D^n \to D^n$.
+  So we now show no such continuous retraction can exist. Suppose $r: D^n \to \partial D^n$ is a retraction, i.e.\ $r \circ i \simeq \id: \partial D^n \to \partial D^n$.
   \[
     \begin{tikzcd}
       S^{n - 1} \ar[r, "i"] & D^n \ar[r, "r"] & S^{n - 1}
@@ -4772,7 +4776,19 @@ \subsection{Rational homology, Euler and Lefschetz numbers}
   \]
   In this case, we have not lost any information because there was no torsion part of the homology groups.
 
-  However, for the non-orientable surfaces, we have
+  However, for the non-orientable surfaces, since
+
+   \[
+    H_k(E_n) =
+    \begin{cases}
+      \Z & k = 0\\
+      \Z^{n - 1} \times \Z / 2 & k = 1\\
+      0 & \text{otherwise}
+    \end{cases},
+  \] 
+
+  (exercise) we have that
+
   \[
     H_k(E_n, \Q) =
     \begin{cases}
@@ -4781,6 +4797,7 @@ \subsection{Rational homology, Euler and Lefschetz numbers}
       0 & \text{otherwise}
     \end{cases},
   \]
+
   This time, this is different from the integral coefficient case, where we have an extra $\Z_2$ term in $H_1$.
 \end{eg}
 
@@ -4866,13 +4883,13 @@ \subsection{Rational homology, Euler and Lefschetz numbers}
   There is an exact sequence
   \[
     \begin{tikzcd}
-      0 \ar[r] & B_i(K; Q) \ar[r] & Z_i(K; \Q) \ar[r] & H_i(K; \Q) \ar[r] & 0
+      0 \ar[r] & B_i(K; \Q) \ar[r] & Z_i(K; \Q) \ar[r] & H_i(K; \Q) \ar[r] & 0
     \end{tikzcd}
   \]
   This is since $H_i(K, \Q)$ is defined as the quotient of $Z_i$ over $B_i$. We also have the exact sequence
   \[
     \begin{tikzcd}
-      0 \ar[r] & Z_i(K; Q) \ar[r] & C_i(K; \Q) \ar[r, "d_i"] & B_{i - 1}(K; \Q) \ar[r] & 0
+      0 \ar[r] & Z_i(K; \Q) \ar[r] & C_i(K; \Q) \ar[r, "d_i"] & B_{i - 1}(K; \Q) \ar[r] & 0
     \end{tikzcd}
   \]
   This is true by definition of $B_{i - 1}$ and $Z_i$. Let $f_i^H, f_i^B, f_i^Z, f_i^C$ be the various maps induced by $f$ on the corresponding groups. Then we have
@@ -4937,7 +4954,7 @@ \subsection{Rational homology, Euler and Lefschetz numbers}
 \end{eg}
 
 \begin{eg}
-  Suppose $G$ is a path-connected topological group, i.e.\ $X$ is a group and a topological space, and inverse and multiplication are continuous maps.
+  Suppose $G$ is a path-connected topological group, i.e.\ $G$ is a group and a topological space, and inverse and multiplication are continuous maps.
 
   If $g \not= 1$, then the map
   \begin{align*}

II_M/probability_and_measure.tex

@@ -703,7 +703,7 @@ \subsection{Probability measures}
     \liminf A_n &= \bigcup_n \bigcap_{m \geq n} A_m.
   \end{align*}
 \end{defi}
-To parse these definitions more easily, we can read $\cap$ as ``for all'', and $\cup$ as ``there exits''. For example, we can write
+To parse these definitions more easily, we can read $\cap$ as ``for all'', and $\cup$ as ``there exist''. For example, we can write
 \begin{align*}
   \limsup A_n &= \forall n,\exists m \geq n\text{ such that }A_m\text{ occurs}\\
   &= \{x: \forall n, \exists m \geq n, x \in A_m\}\\
@@ -1030,9 +1030,9 @@ \subsection{Constructing new measures}
       \draw [->] (0, 0) -- (4, 0);
       \draw [->] (0, 0) -- (0, 4);
 
-      \draw [thick, mblue] (0, 0) -- (1.5, 2) -- (2.5, 2) node [draw, fill=white, circle, inner sep = 0, minimum size = 3] {};
+      \draw [thick, mblue] (0, 0) -- (1.5, 2) -- (2.5, 2) node [circ] {};
 
-      \draw [thick, mblue] (2.5, 3) node [circ] {} -- (4, 4);
+      \draw [thick, mblue] (2.5, 3) node [draw, fill=white, circle, inner sep = 0, minimum size = 3] {} -- (4, 4);
     \end{tikzpicture}
   \end{center}
 \end{eg}