# 微分几何学代写Differential Geometry|MATH 143 Stanford University Assignment

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Measure theory is the branch of mathematics that provides a rigorous framework for the study of integration, probability, and other related areas of analysis. The basic idea of measure theory is to extend the concept of length, area, and volume to more abstract spaces.

In measure theory, a measure is a function that assigns a non-negative number to a set, which represents the size or magnitude of that set. The most important measure is the Lebesgue measure, which is used to define the Lebesgue integral.

Lp Spaces:

Lp spaces are a family of function spaces that are used in functional analysis and in probability theory. The Lp space consists of all functions for which the pth power of the absolute value is Lebesgue integrable.

A cycloid is the plane curve traced out by a point on the circumference of a circle as it rolls without slipping along a straight line. Show that, if the straight line is the $x$-axis and the circle has radius $a>0$, the cycloid can be parametrized as
$$\gamma(t)=a(t-\sin t, 1-\cos t)$$

To show that $\gamma(t)=a(t-\sin t, 1-\cos t)$ parametrizes the cycloid, we need to show that the curve traced out by this parametrization is the same as the curve traced out by a point on the circumference of a circle of radius $a$ as it rolls without slipping along the $x$-axis.

Let $P$ be a point on the circumference of the circle of radius $a$ centered at the origin, and let $Q$ be the point on the $x$-axis directly below $P$, as shown in the figure below.

Let $O$ be the center of the circle, and let $t$ be the angle $\angle AOP$, where $A$ is the point on the $x$-axis directly to the left of $P$. Then the angle $\angle AOQ$ is also $t$, since $OA$ and $OQ$ are radii of the circle.

As the circle rolls without slipping along the $x$-axis, the point $P$ moves along a distance equal to the arc length $s$ from $A$ to $P$. This arc length is equal to the length of the circumference of the circle times the angle $t$ in radians, or $s=at$. Meanwhile, $Q$ moves a distance equal to the horizontal component of $s$, which is $s-\sin t$. Thus, the position of $Q$ at time $t$ is $(s-\sin t, 0)=(at-a\sin t, 0)$. Finally, $P$ and $Q$ are vertically separated by the vertical component of $s$, which is $a(1-\cos t)$. Thus, the position of $P$ at time $t$ is $(at-a\sin t, a(1-\cos t))$.

This shows that the curve traced out by the point $P$ as the circle rolls without slipping along the $x$-axis is given by the parametrization $\gamma(t)=a(t-\sin t, 1-\cos t)$, as desired.

Find the Cartesian equations of the following parametrized curves:
(i) $\gamma(t)=(1+\cos t, \sin t(1+\cos t))$.
(ii) $\gamma(t)=\left(t^2+t^3, t^3+t^4\right)$

(i) To find the Cartesian equations of $\gamma(t)=(1+\cos t, \sin t(1+\cos t))$, we can eliminate the parameter $t$ by solving for $\cos t$ and $\sin t$ in terms of $x$ and $y$.

From the first coordinate, we have $x=1+\cos t$, so $\cos t=x-1$.

From the second coordinate, we have $y=\sin t(1+\cos t)$, so $\sin t=\frac{y}{1+\cos t}=\frac{y}{x}$. Substituting for $\cos t$ and $\sin t$ in terms of $x$ and $y$, we get

\begin{aligned} & x=1+\cos t=1+(x-1)=x, \ & y=\sin t(1+\cos t)=\frac{y}{x} \cdot x(x-1)=y(x-1) .\end{aligned}

Thus, the Cartesian equations of the curve are $x=1+\cos t$ and $y=\sin t(1+\cos t)$, or equivalently,

$x=1+\frac{x^2-y^2}{2 x}, \quad y=\frac{y}{x} \cdot\left(1+\frac{x^2-y^2}{2 x}\right)$.

Simplifying, we get

$x^2-2 x+y^2=1$,

which is the equation of a circle centered at $(1,0)$ with radius $1$.

(ii) To find the Cartesian equations of $\gamma(t)=(t^2+t^3, t^3+t^4)$, we can eliminate the parameter $t$ by solving for $t^2$ and $t^3$ in terms of $x$ and $y$.

From the first coordinate, we have $x=t^2+t^3=t^2(1+t)$, so $t^2=\frac{x}{1+t}$.

From the second coordinate, we have $y=t^3+t^4=t^3(1+t)=\left(\frac{x}{1+t}\right)^{\frac{3}{2}}(1+t)$. Substituting for $t^2$ and $t^3$ in terms of $x$ and $y$, we get

\begin{aligned} t^2 & =\frac{x}{1+t} \ t^3 & =\frac{y}{(1+t)^{\frac{3}{2}}}\end{aligned}

Solving for $t$ in terms of $x$ and $y$, we find

$t=\frac{\sqrt{x^2+4 y}-x}{2}$

Substituting this expression for $t$ into the equation for $x$, we get

$\frac{x}{1+\frac{\sqrt{x^2+4 y}-x}{2}}=\frac{x^2+2 \sqrt{x^2+4 y}-2 x}{\sqrt{x^2+4 y}}=\frac{\left(x^2+2 \sqrt{x^2+4 y}+4 y\right)-4 y}{\sqrt{x^2+4 y}}=\sqrt{x^2+4 y}$.

Calculate the length of the part of the curve
$$\gamma(t)=(\sinh t-t, 3-\cosh t)$$
cut off by the $x$-axis.

To find the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis, we need to find the $t$-values where $\gamma(t)$ intersects the $x$-axis, and then integrate the magnitude of the derivative of $\gamma(t)$ over the interval of $t$ values corresponding to that part of the curve.

The $x$-coordinate of a point on the $x$-axis is zero, so we need to solve the equation $\sinh t – t = 0$ to find the $t$-values where $\gamma(t)$ intersects the $x$-axis. This equation cannot be solved analytically, so we will use numerical methods to find an approximation.

One method is to use the bisection method: we start with an interval that contains a root of the equation (for example, $t \in [1,2]$), and we bisect the interval repeatedly, discarding the half that does not contain a root. After several iterations, we get an interval that contains the root with the desired accuracy.

Using this method, we find that the equation $\sinh t – t = 0$ has a root in the interval $[1.5,1.6]$, which we will denote by $t_0$. To find the length of the part of the curve cut off by the $x$-axis, we need to integrate the magnitude of the derivative of $\gamma(t)$ over the interval $[0,t_0]$.

The derivative of $\gamma(t)$ is

$\gamma^{\prime}(t)=(\cosh t-1, \sinh t)$

and its magnitude is

$\left|\gamma^{\prime}(t)\right|=\sqrt{(\cosh t-1)^2+(\sinh t)^2}=\sqrt{\cosh ^2 t-2 \cosh t+2}$

To integrate this expression over the interval $[0,t_0]$, we make the substitution $u = \sinh t$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} \sqrt{\cosh ^2\left(\sinh ^{-1} u\right)-2 \cosh \left(\sinh ^{-1} u\right)+2} d u$.

Using the identity $\cosh^2 x – \sinh^2 x = 1$, we can simplify the expression under the square root to $2 – 2 \sinh(\sinh^{-1} u) = 2(1 – u^2)$, so that

$\int_0^{t_0}\left|\gamma^{\prime}(t)\right| d t=\int_0^{\sinh t_0} 2 \sqrt{1-u^2} d u=\pi$

Therefore, the length of the part of the curve $\gamma(t)$ cut off by the $x$-axis is $\boxed{\pi}$.

# 随机过程代写Stochastic Processes|MATH 136 STATS 219 Stanford University Assignment

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A graph $G$ is connected when, for two vertices $x$ and $y$ of $G$, there exists a sequence of vertices $x_0, x_1, \ldots, x_k$ such that $x_0=x, x_k=y$, and $x_i \sim x_{i+1}$ for $0 \leq i \leq k-1$. Show that random walk on $G$ is irreducible if and only if $G$ is connected.

Proof. Let $P$ denote the transition matrix of random walk on $G$. The random walk is irreducible if for any vertices $x$ and $y$ there exists an integer $k$ such that $P^k(x, y)>0$. Note that $P^k(x, y)>0$ if and only if there exist vertices $x_0=x, x_1, \ldots, x_k=y$ such that $\prod_{i=0}^{k-1} P\left(x_i, x_{i+1}\right)>0$, i.e., $x_i \sim x_{i+1}$ for all $0 \leq i \leq k-1$. Therefore, the random walk is irreducible if and only if $G$ is connected.

We define a graph to be a tree if it is connected but contains no cycles. Prove that the following statements about a graph $T$ with $n$ vertices and $m$ edges are equivalent:
(a) $T$ is a tree.

Proof. The equivalence can be easily seen from Euler’s formula $m=n+l-2$ where $l$ denotes the number of faces of the graph, because any two of the following conditions will imply the other:
(1) $T$ is connected $\Longleftrightarrow m=n+l-2$;
(2) $T$ has no cycles $\Longleftrightarrow l=1$;
(3) $m=n-1$
Since this simple equivalence is a special case (and sometimes the starting point of the proof) of Euler’s formula, it should be proved without the use of the more general theorem. We provide a long yet elementary proof here. All three parts of the following proof are based on a simple operation, namely, removing one edge and one vertex at a time. We assume without loss of generality that $G$ has at least one edge. First we need a claim.
Claim: If each vertex of a graph $G$ has degree at least 2, then $G$ contains a cycle.
Start from any vertex $x_0$ of $G$ and we can find $x_1 \sim x_0$. Suppose we already find distinct $x_0, \ldots, x_i$ such that $x_0 \sim x_1 \sim \cdots \sim x_i$. Since $x_i$ has degree at least 2 , we can find $x_{i+1} \neq x_{i-1}$ such that $x_i \sim x_{i+1}$. If $x_{i+1}=x_j$ for some $j<i-1$, then we form a cycle. Otherwise we continue the process. The process must end because $G$ is finite, so $G$ contains a cycle.
(a) implies $(b)$ : Since $T$ is connected and contains no cycles, the claim implies that there exists a vertex of degree 1 in $T$. We delete this vertex and the attached edge from $T$, and the remaining object $T^{\prime}$ is still a connected graph with no cycles. We continue this process until the remaining graph has only one edge and thus two vertices. Since at each step we delete one edge and one vertex, it follows that $m=n-1$.

(b) $T$ is connected and $m=n-1$.
(c) $T$ has no cycles and $m=n-1$.

(b) implies (c): If there exists a vertex of degree 1 in $T$, we delete this vertex and the attached edge from $T$. Then the remaining object $T^{\prime}$ is still a connected graph with $m^{\prime}=n^{\prime}-1$ where $m^{\prime}$ is the number of edges and $n^{\prime}$ is the number of vertices. We continue this process until the remaining graph has no edges, or every vertex has degree at least 2 . The second case cannot happen because otherwise $n^{\prime} \leq m^{\prime}$ which is a contradiction. In the first case, $T$ cannot contain a cycle, because otherwise when we first delete an edge and one of its vertex in a cycle, the remaining object is no longer a graph.
(c) implies (a): If there exists a vertex of degree 1 in $T$, we delete this vertex and the attached edge from $T$. The remaining object $T^{\prime}$ is still a graph with no cycles and $m^{\prime}=n^{\prime}-1$. Note that if $T$ is not connected, then $T^{\prime}$ is not connected. We continue this process until the remaining graph has no edges, or every vertex has degree at least 2. The second case contradicts the claim because $T$ has no cycles. In the first case, because the relation $m=n-1$ is preserved, the remaining graph contains exactly one vertex and is thus connected. We conclude that $T$ is connected.

# 代数几何学代写Algebraic Geometry|MATH 145 Stanford University Assignment

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Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 145 Algebraic Geometry代数几何学代写代考辅导服务！

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Algebraic geometry is a branch of mathematics that studies the properties of algebraic varieties, which are geometric objects defined by polynomial equations. The subject combines techniques from algebra and geometry to study the solutions of polynomial equations and their geometric properties.

In algebraic geometry, the main objects of study are algebraic varieties, which are geometric objects defined by polynomial equations. For example, the equation x^2 + y^2 = 1 defines a circle in the plane, and the equation x^3 + y^3 = 1 defines a curve known as the twisted cubic. Algebraic varieties can be defined over any field, but in classical algebraic geometry the focus is on varieties defined over the complex numbers.

One of the key tools in algebraic geometry is the notion of a scheme, which generalizes the notion of an algebraic variety to include more general types of objects. Schemes are studied using sheaf theory, which provides a way to associate to each point of a scheme a local ring, which encodes information about the geometry of the scheme near that point.

Algebraic geometry has a wide range of applications, including in number theory, cryptography, physics, and computer science. It has also been influential in other areas of mathematics, such as topology and representation theory. Some of the most famous results in algebraic geometry include the proof of Fermat’s Last Theorem by Andrew Wiles, and the proof of the Hodge Conjecture by Pierre Deligne.

(a) Prove that if $X=\operatorname{Spec}(A)$ is affine and locally factorial, then $P i c(X)$ is trivial iff $A$ is a UFD.

First, suppose that $Pic(X)$ is trivial. Since $X$ is affine and locally factorial, the divisor class group $Cl(X)$ is also trivial. In other words, every Weil divisor on $X$ is linearly equivalent to a principal divisor. Let $f$ be a nonzero element of $A$, and consider the principal divisor $(f)$. Since $Pic(X)$ is trivial, $(f)$ is the divisor of a regular function $g$ on $X$. This means that $g$ has a pole of order $1$ at every point of $V(f)$ (the vanishing locus of $f$) and is regular elsewhere. In particular, $g$ is regular on the complement of $V(f)$, which is the open set $D(f)$. Therefore, $g$ is a unit in $A_f$, the localization of $A$ at $f$. This implies that $(f)$ is the trivial divisor, since it is linearly equivalent to the zero divisor $(1)$.

Now, suppose that $A$ is a UFD. Let $D$ be a Weil divisor on $X$, given by a nonzero element $f$ of $A$ and a nonzero function $g$ on $D$. We can write $g$ as $g=au/f^n$ for some unit $u$ of $A$, some integer $n\geq 0$, and some element $a$ of $A$ not divisible by $f$. We define a principal divisor $(g)$ as the formal sum $\sum_{x\in X} v_x(g) \cdot [x]$, where $v_x(g)$ denotes the order of vanishing of $g$ at $x$ and $[x]$ denotes the corresponding point of $X$. Note that $v_x(g)$ is zero for all but finitely many points of $X$, since $g$ is nonzero. Also note that $v_x(g)$ is equal to $n$ if $x\in D$ and $v_x(f)$ if $x\not\in D$. Therefore, $(g)$ is linearly equivalent to the divisor $(f^{n-v_x(g)})$.

Let $h$ be another nonzero function on $D$, represented by a nonzero element $b$ of $A$ and a unit $v$ of $A$. We have $h=bv/f^m$ for some integer $m\geq 0$. The ratio $g/h$ is then equal to $u(v/b)(f^{m-n})$, which is a unit of $A$ (since $A$ is a UFD). Therefore, $(g)$ and $(h)$ are linearly equivalent as divisors on $X$. This shows that every Weil divisor on $X$ is linearly equivalent to a principal divisor, so $Cl(X)$ is trivial. Since $Pic(X)$ is isomorphic to $Cl(X)$ (as explained, for example, in Hartshorne’s Algebraic Geometry, Chapter II, Section 6), we conclude that $Pic(X)$ is also trivial.

(b) Let $X \subset \mathbb{P}^n$ be a projective variety. Suppose that the homogeneous coordinate ring of $X$ is a UFD. Show that $\operatorname{Pic}(X) \cong \mathbb{Z}$.

Since $X$ is a projective variety, it is a closed subset of $\mathbb{P}^n$ for some $n$, and hence we may assume that $X$ is a closed subset of $\mathbb{P}^n$ for some $n$. Let $S(X)$ be the homogeneous coordinate ring of $X$, which is a graded ring.

Since $S(X)$ is a UFD, every irreducible homogeneous element of $S(X)$ is a prime element. In particular, every irreducible hypersurface in $\mathbb{P}^n$ that intersects $X$ is a prime divisor on $X$.

Conversely, every prime divisor $D$ on $X$ corresponds to an irreducible hypersurface in $\mathbb{P}^n$ that intersects $X$. By the projective Nullstellensatz, the homogeneous ideal of this hypersurface is generated by homogeneous polynomials of the same degree, and we may assume that these polynomials are irreducible. Since $S(X)$ is a UFD, the irreducible polynomials generate a prime ideal, and hence $D$ is an irreducible element of $\operatorname{Pic}(X)$.

Thus, we have established a bijection between the prime divisors on $X$ and the irreducible elements of $\operatorname{Pic}(X)$. Moreover, every non-zero element of $\operatorname{Pic}(X)$ can be written as a product of irreducible elements, since $S(X)$ is a UFD.

Therefore, $\operatorname{Pic}(X)$ is isomorphic to the group of divisors modulo principal divisors, which is generated by the class of any prime divisor on $X$. Since $X$ is irreducible, any non-empty open subset of $X$ is dense in $X$, and hence any non-zero regular function on $X$ is a unit in $S(X)$. It follows that the group of principal divisors on $X$ is isomorphic to $\mathbb{Z}$, and hence $\operatorname{Pic}(X) \cong \mathbb{Z}$.

Let $X$ be the line with a double point at zero, thus we have a map $X \rightarrow \mathbb{A}^1$ which is an isomorphism over $\mathbb{A}^1 \backslash{0}$ and the preimage of 0 consists of two points.
(a) Let $Y=\mathbb{A}^2 \backslash{0}$. Show that the map $m: Y \rightarrow \mathbb{A}^1, m(x, y)=x y$ can be lifted to an onto map $Y \rightarrow X$; moreover, there are two distinct such liftings.

To construct a lifting, we need to find a map $f: Y \rightarrow X$ such that $m = \pi \circ f$, where $\pi: X \rightarrow \mathbb{A}^1$ is the given projection map. Since $m(x,y) = xy$ and $\pi$ has a double point at zero, it’s natural to define $f$ as follows:

• For $(x,y) \in Y$ with $xy \neq 0$, let $f(x,y) = (x,y)$, i.e., we map $(x,y)$ to itself.
• For $(x,y) \in Y$ with $xy = 0$, let $f(x,y) = (x,0)$ if $y=0$, and $f(x,y) = (0,y)$ if $x=0$.

It’s clear that $f$ is well-defined and continuous on $Y$. Moreover, since $f(x,y) = (x,y)$ for $(x,y) \in Y \cap (\mathbb{A}^1 \times \mathbb{A}^1)$, where $\mathbb{A}^1 \times \mathbb{A}^1$ is the open subset of $Y$ where both coordinates are nonzero, we have $\pi \circ f = m$ on this open subset. Therefore, $\pi \circ f = m$ on all of $Y$, since both $m$ and $\pi \circ f$ are continuous.

To show that there are two distinct liftings, note that we can also define $f’: Y \rightarrow X$ by interchanging the two coordinates in the second case above:

• For $(x,y) \in Y$ with $xy \neq 0$, let $f'(x,y) = (x,y)$, i.e., we map $(x,y)$ to itself.
• For $(x,y) \in Y$ with $xy = 0$, let $f'(x,y) = (0,y)$ if $x=0$, and $f'(x,y) = (x,0)$ if $y=0$.

Then $f’$ is also well-defined and continuous on $Y$, and satisfies $\pi \circ f’ = m$. However, $f$ and $f’$ are not equal, since $f(1,0) = (1,0)$ and $f'(1,0) = (0,0)$. Therefore, we have found two distinct liftings of $m$ to $X$.

# 拓扑学和几何学代写Introduction to Topology and Geometry|MATH 144 Stanford University Assignment

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Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 144 ntroduction to Topology and Geometry拓扑学和几何学代写代考辅导服务！

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Topology and geometry are two branches of mathematics that deal with the study of shapes and spaces. While they share some similarities, they also have important differences.

Topology is the branch of mathematics concerned with the properties of space that are preserved under continuous transformations, such as stretching or bending. In topology, the focus is on the study of the properties of space that are independent of the exact shape or size of the space. For example, a circle and a square may look very different, but they share important topological properties, such as being closed curves without endpoints.

Geometry, on the other hand, is the branch of mathematics concerned with the study of shape, size, and position of objects in space. Unlike topology, geometry is concerned with the exact properties of space, such as angles, distances, and shapes. Geometry is traditionally divided into two main branches: plane geometry, which deals with objects in two-dimensional space, and solid geometry, which deals with objects in three-dimensional space.

While topology and geometry have different focuses, they are closely related. Many geometric properties can be studied using topological methods, and topology can provide a useful framework for understanding the underlying structure of geometric objects. As a result, topology and geometry often intersect and overlap in interesting and useful ways.

(a) Let $[n]$ denote the totally ordered set $\{0,1, \ldots, n\}$. Let $\phi:[m] \rightarrow[n]$ be an order preserving function (so that if $i \leq j$ then $\phi(i) \leq \phi(j)$ ). Identifying the elements of $[n]$ with the vertices of the standard simplex $\Delta^n, \phi$ extends to an affine map $\Delta^m \rightarrow \Delta^n$ that we also denote by $\phi$. Give a formula for this map in terms of barycentric coordinates: If $\phi\left(s_0, \ldots, s_m\right)=\left(t_0, \ldots, t_n\right)$, what is $t_j$ as a function of $\left(s_0, \ldots, s_m\right) ?$

We can express the vertices of the standard simplex $\Delta^n$ as the barycentric coordinates $(0,0,\ldots,0,1)$, $(0,0,\ldots,1,0)$, $(0,0,\ldots,1)$, $(0,0,\ldots,1,1,0)$, $(0,0,\ldots,1,0,1)$, $\ldots$, $(1,0,\ldots,0,0)$, $(0,1,\ldots,0,0)$, $(0,0,\ldots,0,1)$, where the $1$ appears in the $i$-th position.

Let $\phi:[m] \rightarrow[n]$ be an order-preserving function, and suppose that $\phi(s_i) = t_j$. Then the affine map $\phi:\Delta^m \rightarrow \Delta^n$ sends the barycentric coordinate $(s_0,\ldots,s_m)$ to the point in $\Delta^n$ whose barycentric coordinates are $(t_0,\ldots,t_n)$.

To compute $t_j$, we note that the $j$-th coordinate of the barycentric coordinates of a point in $\Delta^n$ is given by the sum of the weights of the vertices of $\Delta^n$ that lie in the $j$-th face of $\Delta^n$. In other words, $t_j$ is the sum of the weights of the vertices of $\Delta^n$ whose $j$-th coordinate is $1$.

Let $I_j$ denote the set of indices $i$ such that $\phi(s_i) = j$. Then we have \begin{align*} t_j &= \sum_{i \in I_j} w_i \ &= \sum_{i \in I_j} \prod_{k \neq i} \frac{s_k}{s_k – s_i}, \end{align*} where $w_i$ denotes the weight of the vertex of $\Delta^m$ whose $i$-th coordinate is $1$. The second equality follows from the formula for barycentric coordinates of a point in the simplex.

(b) Write $d^j:[n-1] \rightarrow[n]$ for the order preserving injection that omits $j$ as a value. Show that an order preserving injection $\phi:[n-k] \rightarrow[n]$ is uniquely a composition of the form $d^{j_k} d^{j_{k-1}} \cdots d^{j_1}$, with $0 \leq j_1<j_2<\cdots<j_k \leq n$. Do this by describing the integers $j_1, \ldots, j_k$ directly in terms of $\phi$, and then verify the straightening rule
$$d^i d^j=d^{j+1} d^i \quad \text { for } \quad i \leq j$$

To describe the integers $j_1,\ldots,j_k$ directly in terms of $\phi$, we proceed by induction on $k$. For $k=1$, we let $j_1$ be the unique value $j$ such that $\phi$ maps $[j-1]$ bijectively onto $\phi([j-1])$ and maps $j$ to $\phi([j-1])\cup{\phi(j)}$. This value exists because $\phi$ is order preserving, so it maps each element of $[n-k]$ to an interval contained in $[n]$, and hence maps $n-k$ to an interval contained in $[n-1]$. By the induction hypothesis, we may assume that $\phi=d^{j_1}\circ\cdots\circ d^{j_{k-1}}\circ d^{j_k’}$ for some integers $j_1<\cdots<j_{k-1}<j_k’\leq n-1$. Let $j_k$ be the unique value such that $\phi$ maps $[j_k-1]$ bijectively onto $\phi([j_k-1])$ and maps $j_k$ to $\phi([j_k-1])\cup{\phi(j_k’)}$. Again, this value exists because $\phi$ is order preserving and maps $j_k’$ to an interval contained in $[n]$. By construction, we have $j_k’>j_{k-1}$, so $j_k$ is strictly greater than all the integers $j_1,\ldots,j_{k-1}$.

We now prove the straightening rule by induction on $i$. For the base case $i=0$, note that $d^0$ is the identity map and hence commutes with $d^j$ for any $j$. For the induction step, suppose that $i\geq 1$ and let $\phi=d^{j_k}\circ\cdots\circ d^{j_1}$ and $\psi=d^i\circ d^j$, where $0\leq j\leq i\leq n-1$. We need to show that $\psi$ can be expressed in the form $d^{j_k’}\circ\cdots\circ d^{j_1′}\circ d^{j_{k+1}’}\circ d^{j_k}\circ\cdots\circ d^{j_{i+1}}$, where $j_1′,\ldots,j_{k+1}’$ are integers such that $0\leq j_1′<\cdots<j_{k+1}’\leq n$. We consider two cases.

Case 1: $j<i\leq j+1$. In this case, we have $\psi=d^i\circ d^j=d^{j+1}\circ d^i$, so we can take $j_1′,\ldots,j_{k+1}’$ to be the same as $j_1,\ldots,j_k,j+1,i$ in some order.