nyuuzyou/wikis huggingface dataset:
Homotopic Chain Maps Induce Equal Maps on Homology

== Theorem ==
Let $A_\bullet$, $B_\bullet$ be chain complexes of abelian groups.
Let $f, g: A_\bullet \to B_\bullet$ be chain maps which are homotopic.
Then $f$ and $g$ induce equal maps on homology.
== Proof ==
Let $\partial^A_\bullet, \partial^B_\bullet$ be the differentials on $A_\bullet$ and $B_{\bullet}$ respectively.
Let $h$ be a homotopy between $f$ and $g$. 
Let:
:$a \in \map {H_n} A \cong \map \ker {\partial^A_n} / \Img {\partial^A_{n + 1} }$
There exists $\tilde a \in \map \ker {\partial^A_n}$ representing $a$.
It is enough to show that:
:$\map {f_n} {\tilde a} \sim \map {g_n} {\tilde a}$
Equivalently:
:$\map {f_n} {\tilde a} - \map {g_n} {\tilde a} \in \Img {\partial^B_{n + 1} }$
We know that:
:$\partial h - h \partial = f - g$
This means:
:$\partial^B_{n + 1} h_n - h_{n - 1} \partial^A_n = f_n - g_n$
Plugging in $\tilde a$: 
:$\map {\partial^B_{n + 1} } {\map {h_n} {\tilde a} } - \map {h_{n - 1}} {\map {\partial^A_n} {\tilde a} } = \map {f_n} {\tilde a} - \map {g_n} {\tilde a}$
Since $\tilde a \in \map \ker {\partial^A_n}$:
:$\map {\partial^B_{n + 1} } {\map {h_n} {\tilde a} } = \map {f_n} {\tilde a} - \map {g_n} {\tilde a}$
Therefore:
:$\map {f_n} {\tilde a} - \map {g_n} {\tilde a} \in \Img {\partial^B_{n + 1} }$
Category:Homological Algebra
Homotopic Paths Implies Homotopic Composition

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $f_1, f_2, g_1, g_2: \left[{0 \,.\,.\, 1}\right] \to S$ be paths in $T$.
Let $f_1$ be homotopic to $f_2$ and $g_1$ be homotopic to $g_2$.
Then the concatenated paths $f_1 * g_1$ and $f_2 * g_2$ are homotopic.
== Proof ==
Let $F: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $f_1$ and $f_2$.
Let $G: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $g_1$ and $g_2$.
Define $H: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ by:
:$H \left({s, t}\right) = \begin{cases} F \left({2 s, t}\right) & : s \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\
G \left({2 s - 1, t}\right) & : s \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]\end{cases}$
By Continuous Mapping on Finite Union of Closed Sets, $H$ is continuous.
By definition of concatenation of paths, $H$ is a homotopy between $f_1 * g_1$ and $f_2 * g_2$.
== Also see ==
* Definition:Multiplication of Homotopy Classes of Paths
== Sources ==
* : $9$: The Fundamental Group: $\S 52$: The Fundamental Group
Category:Homotopy Theory
Homotopic Paths have Same Endpoints

== Definition ==
Let $X$ be a topological space.
Let $f, g: \closedint 0 1 \to X$ be paths.
Let $f$ and $g$ be homotopic.
Then $f$ and $g$ have the same endpoints.
That is:
:$\map f 0 = \map g 0$ and $\map f 1 = \map g 1$.
== Proof ==
By , $f$ and $g$ are homotopic relative to $\set {0, 1}$.
By :
:$\forall x \in \set {0, 1}: \map f x = \map g x$
The result follows.
Category:Homotopy Theory
Homotopy-Class of Curves in Complete Connected Riemannian Manifold contains Minimal Geodesic of Class

== Theorem ==
Let $\struct {M, g}$ be a complete connected Riemannian manifold.
Let $H_{p q}$ homotopy class of segments from $p \in M$ to $q \in M$.
Let $\gamma_{p q}$ be a geodesic from $p \in M$ to $q \in M$ such that $\gamma_{p q} \in H_{p q}$.
Let $y_{p q} \in H_{p q}$ be an admissible curve.
Let $L_g$ be the Riemannian length.
Then in every $H_{p q}$ there is a $\gamma_{pq}$ which is minimizing among all $y_{p q} \in H_{p q}$:
:$\forall p, q \in M : \exists \gamma_{pq} \in H_{p q} : \forall y_{p q} \in H_{p q} : \map {L_g} {\gamma_{p q} } \le \map {L_g} {y_{p q} }$
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Homotopy Group is Homeomorphism Invariant

== Theorem ==
Let $X$ and $Y$ be two topological spaces.
Let $\phi: X \to Y$ be a homeomorphism.
Let $x_0 \in X$, $y_0 \in Y$.
Then for all $n \in \N$ the induced mapping:
:$\phi_* : \pi_n \left({X, x_0}\right) \to \pi_n \left({Y, y_0}\right):$
::$\left[{\!\left[{\, c \,}\right]\!}\right] \mapsto \left[{\!\left[{\, \phi \circ c \,}\right]\!}\right]$
is an isomorphism, where $\pi_n$ denotes the $n$th homotopy group.
== Proof ==
Let $\phi: X \to Y$ be a homeomorphism.
We must show that:
:$(1): \quad$ If $c: \left[{0 \,.\,.\, 1}\right]^n \to X$ is a continuous mapping, then $ \phi \circ c: \left[{0 \,.\,.\, 1}\right]^n \to Y$ is also continuous
:$(2): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are freely homotopic, then $\phi \circ c, \phi \circ d: \left[{0 \,.\,.\, 1}\right]^n \to Y$ are also freely homotopic
:$(3): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are not freely homotopic, there can be no free homotopy between $\phi \circ c$ and $\phi \circ d$
:$(4): \quad$ The image of the concatenation of two maps, $\phi \left({c * d}\right)$, is freely homotopic to the concatenation of the images, $\phi \left({c}\right) * \phi \left({d}\right)$.
Category:Homotopy Theory
Category:Algebraic Topology
Honsberger's Identity

== Theorem ==
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
=== Negative Indices ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: $(6)$
Category:Fibonacci Numbers
Category:Honsberger's Identity
Honsberger's Identity/Negative Indices

== Theorem ==
Let $n \in \Z_{< 0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then Honsberger's Identity:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
continues to hold, whether $m$ or $n$ are positive or negative.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
This can equivalently be written:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$F_{m - n} = F_{m - 1} F_{-n} + F_m F_{-n + 1}$
$\map P 0$ is the case:
Thus $\map P 0)$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ and $\map P {k - 1}$ are true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$F_{m - k} = F_{m - 1} F_{-k} + F_m F_{-k + 1}$
and:
:$F_{m - \paren {k - 1} } = F_{m - 1} F_{-\paren {k - 1} } + F_m F_{-\paren {k - 1} + 1}$
from which it is to be shown that:
=== Induction Step ===
First we note that:
and:
This is the induction step:
So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\le 0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
</onlyinclude>
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $9$
Category:Honsberger's Identity
Honsberger's Identity/Proof 1

== Theorem ==
== Proof ==
From the initial definition of Fibonacci numbers, we have:
:$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$
Proof by induction:
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\forall m \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ is seen to hold.
$\map P 2$ is the case:
and so $\map P 2$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ and $\map P {k - 1}$ are true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$F_{m + k} = F_{m - 1} F_k + F_m F_{k + 1}$
and:
:$F_{m + k - 1} = F_{m - 1} F_{k - 1} + F_m F_k$
from which it is to be shown:
:$F_{m + k + 1} = F_{m - 1} F_{k + 1} + F_m F_{k + 2}$
=== Induction Step ===
This is our induction step:
So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Category:Honsberger's Identity
Category:Proofs by Induction
Hooke's Law

== Physical Law ==
Hooke's Law applies to an ideal spring:
:$\mathbf F = -k \mathbf x$
where:
:$\mathbf F$ is the force caused by a displacement $\mathbf x$
:$k$ is the spring force constant.
The negative sign indicates that the force pulls in the opposite direction to that of the displacement imposed.
:The strain is proportional to the stress.
=== Application to Physical Body ===
== Also see ==
* Dimension of Spring Force Constant: the dimension of $k$ is $\mathsf {M T}^{-2}$.
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: The Gravitational Constant
* 
* 
* 
* 
* 
* 
Category:Hooke's Law
Category:Elasticity
Category:Physics
Hopf-Rinow Theorem/Corollary 1

== Theorem ==
Let $\struct {M, g}$ be a connected Riemannian manifold.
Let $T_p M$ be the tangent space at $p \in M$.
Let $\exp_p$ be the restricted exponential map.
Suppose there is $p \in M$ such that $\exp_p$ is defined on all of $T_p M$.
Then $M$ is complete.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Hopf-Rinow Theorem/Corollary 2

== Theorem ==
Let $\struct {M, g}$ be a complete connected Riemannian Manifold.
Let $\gamma$ be a minimizing geodesic segment.
Then any two points of $M$ can be joined by some $\gamma$:
:$\forall p, q \in M : \exists \gamma : \paren {p \in \gamma} \land \paren{q \in \gamma}$
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Hopf-Rinow Theorem/Corollary 3

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian Manifold.
Let $\gamma$ be a maximal geodesic in $M$.
Then:
:$\Dom \gamma = \R$ 
where $\Dom \gamma$ denotes the domain of $\gamma$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Horizontal Point of Inflection is Stationary Point

== Theorem ==
Let $f$ be a real function which is differentiable on an interval $\Bbb I \subseteq \R$.
Let $\xi \in \Bbb I$ be such that $\xi$ has a point of inflection at $\xi$ such that the tangent to $f$ at $\xi$ is parallel to the $x$-axis.
Then $\xi$ is a stationary point.
== Proof ==
== Sources ==
* 
* 
Category:Points of Inflection
Category:Stationary Points
Horizontal Section of Cartesian Product

== Theorem ==
Let $X$ and $Y$ be sets.
Let $A \subseteq X$ and $B \subseteq Y$, so that $A \times B \subseteq X \times Y$. 
Let $y \in Y$. 
Then: 
:$\paren {A \times B}^y = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$
where $\paren {A \times B}^y$ is a horizontal section of $A \times B$.
== Proof ==
Let $y \in B$. 
From the definition of the horizontal section, we have: 
:$x \in \paren {A \times B}^y$
:
:$\tuple {x, y} \in A \times B$ 
Since $y \in B$, this equivalent to: 
:$x \in A$
So:
:$x \in \paren {A \times B}^y$ $x \in A$
giving:
:$\paren {A \times B}^y = A$ if $y \in B$.
Now let $y \in Y \setminus B$.
So, by the definition of set difference, we have $y \in Y$ and $y \not \in B$.
As before, we have:
:$x \in \paren {A \times B}^y$ 
:
:$\tuple {x, y} \in A \times B$
But this is equivalent to: 
:$x \in A$ and $y \in B$.
Since $y \not \in B$, there exists no $x \in \paren {A \times B}^y$.
So:
:$\paren {A \times B}^y = \O$ if $y \not \in B$.
Category:Horizontal Section of Sets
Category:Cartesian Product
Horizontal Section of Characteristic Function is Characteristic Function of Horizontal Section

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $E \subseteq X \times Y$.
Let $y \in Y$. 
Then: 
:$\paren {\chi_E}^y = \chi_{E^y}$
where:
:$E^y$ is the $y$-horizontal section of $E$
:$\chi_E$ and $\chi_{E^y}$ are the characteristic functions of $E$ and $E^y$ as subsets of $X \times Y$ respectively
:$\paren {\chi_E}^y$ is the $y$-horizontal function of $\chi_E$.
== Proof ==
We show that: 
:$\map {\paren {\chi_E}^y} x = \begin{cases}1 & x \in E^y \\ 0 & x \not \in E^y\end{cases}$
at which point we'll be done from the definition of the characteristic function of $E^y$. 
From the definition of the $y$-horizontal section, we have: 
:$\map {\paren {\chi_E}^y} x = \map {\chi_E} {x, y}$ 
From the definition of a characteristic function, we have: 
:$\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \in E$.
From the definition of the $y$-horizontal section, we then have: 
:$\map {\chi_E} {x, y} = 1$ $x \in E^y$. 
So:
:$\map {\paren {\chi_E}^y} x = 1$ $x \in E^y$. 
From the definition of a characteristic function, we also have: 
:$\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \not \in E$.
From the definition of the $y$-horizontal section, we then have: 
:$\map {\chi_E} {x, y} = 0$ $x \not \in E^y$.
So:
:$\map {\paren {\chi_E}^y} x = 0$ $x \not \in E^y$
giving:
:$\map {\paren {\chi_E}^y} x = \begin{cases}1 & x \in E^y \\ 0 & x \not \in E^y\end{cases}$
which was the demand.
Category:Characteristic Functions
Category:Horizontal Section of Functions
Horizontal Section of Continuous Function is Continuous

== Theorem ==
Let $X$, $Y$ and $T$ be topological spaces.
Equip the Cartesian product $X \times Y$ with the product topology.
Let $f : X \times Y \to T$ be a continuous mapping. 
Let $y \in Y$.
Then the $y$-horizontal section $f^y : X \to T$ is continuous.
== Proof ==
From the definition of the $y$-horizontal section, we have: 
:$\map {f^y} x = \map f {x, y}$
for each $x \in X$. 
Define the map $p^y : X \to X \times Y$ by: 
:$\map {p^y} x = \tuple {x, y}$ 
for each $x \in X$. 
We have that: 
:$f^y = f \circ p^y$
From Composite of Continuous Mappings is Continuous, since $f$ is continuous, it suffices to show that $p^y$ is continuous.
From Box Topology on Finite Product Space is Product Topology, the product topology on $X \times Y$ is also the box topology. 
From Continuity Test using Basis, it then suffices to check that: 
:$\paren {p^y}^{-1} \sqbrk {U \times V}$ is open whenever $U$ is open in $X$ and $V$ is open in $Y$.
Let $U$ be open in $X$ and $V$ be open in $Y$.
Note that: 
:$x \in \paren {p^y}^{-1} \sqbrk {U \times V}$
:
:$\tuple {x, y} \in U \times V$
which is equivalent to: 
:$x \in U$ and $y \in V$. 
Clearly then if $y \not \in V$, we have $\paren {p^y}^{-1} \sqbrk {U \times V} = \O$.
We can also read off from this equivalence that if $y \in V$, we have $\paren {p^y}^{-1} \sqbrk {U \times V} = U$.
From Empty Set is Element of Topology, $\O$ is open in $Y$.
Since $U$ is open in $X$ by hypothesis, we therefore have that $p^y$ is continuous from Continuity Test using Basis.
Hence the result.
Category:Horizontal Section of Functions
Category:Continuous Mappings
Category:Product Topology
Horizontal Section of Empty Set

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $y \in Y$. 
Then: 
:$\O^y = \O$
where $\O$ is the empty set and $\O^y$ is the $y$-horizontal section of $\O$. 
== Proof ==
 suppose that: 
:$x \in \O^y$
Then from the definition of the $x$-vertical section, we have: 
:$\tuple {x, y} \in \O$
This is impossible from the definition of the empty set.
So:
:there exists no $x \in \O^y$
giving:
:$\O_x = \O$
Category:Horizontal Section of Sets
Horizontal Section of Linear Combination of Functions is Linear Combination of Horizontal Sections

== Theorem ==
Let $X$ and $Y$ be sets.
Let $f_1, f_2, \ldots, f_n : X \times Y \to \overline \R$ be functions. 
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be real numbers.
Let $y \in Y$. 
Then: 
:$\ds \paren {\sum_{k \mathop = 1}^n \alpha_k f_k}^y = \sum_{k \mathop = 1}^n \alpha_k \paren {f_k}^y$
where $f^y$ denotes the $y$-horizontal section of the function $f$. 
== Proof ==
Let $x \in X$. 
We have:
so:
:$\ds \paren {\sum_{k \mathop = 1}^n \alpha_k f_k}^y = \sum_{k \mathop = 1}^n \alpha_k \paren {f_k}^y$
Category:Horizontal Section of Functions
Horizontal Section of Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.
Let $f : X \times Y \to \overline \R$ be a $\Sigma_X \otimes \Sigma_Y$-measurable function where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$. 
Let $y \in Y$.
Then: 
:$f^y$ is $\Sigma_X$-measurable
where $f^y$ is the $y$-horizontal section of $f$.
== Proof ==
By the definition of a $\Sigma_X$-measurable function, we have that: 
:$f^{-1} \sqbrk D \in \Sigma_X \otimes \Sigma_Y$ for each Borel set $D \subseteq \R$.
We aim to show that: 
:$\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$ for each Borel set $D \subseteq \R$.
Let $D \subseteq \R$ be a Borel set. 
From Preimage of Horizontal Section of Function is Horizontal Section of Preimage, we have: 
:$\paren {f^y}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}^y$
From Horizontal Section of Measurable Set is Measurable, we have: 
:$\paren {f^{-1} \sqbrk D}^y \in \Sigma_X$
so:
:$\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$
So:
:$\paren {f^y}^{-1} \sqbrk D \in \Sigma_X$ for each Borel set $D \subseteq \R$.
so:
:$f^y$ is $\Sigma_X$-measurable.
Category:Horizontal Section of Functions
Category:Measurable Functions
Horizontal Section of Measurable Set is Measurable

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces. 
Let $E \in \Sigma_X \otimes \Sigma_Y$ where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.
Let $y \in Y$. 
Then: 
:$E^y \in \Sigma_X$
where $E^y$ is the $y$-horizontal section of $E$. 
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : $5.1$: Constructions: Lemma $5.1.2$
Category:Horizontal Section of Sets
Category:Horizontal Section of Measurable Set is Measurable
Horizontal Section of Measurable Set is Measurable/Proof 1

== Theorem ==
== Proof ==
Let:
:$\FF = \set {E \subseteq X \times Y : E^y \in \Sigma_X}$
We will show that $\FF$ contains each $S_1 \times S_2$ with $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
We will then show that $\FF$ is a $\sigma$-algebra, at which point we will have: 
:$\map \sigma {\set {S_1 \times S_2 : S_1 \in \Sigma_X, \, S_2 \in \Sigma_Y} } \subseteq \FF$
from Sigma-Algebra Contains Generated Sigma-Algebra of Subset.
From the definition of the product $\sigma$-algebra, we will then have: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$
We will then have the demand.
Let $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
From Horizontal Section of Cartesian Product, we have: 
:$\ds \paren {S_1 \times S_2}^y = \begin{cases}S_1 & y \in S_2 \\ \O & y \not \in S_2\end{cases}$
From the definition of a $\sigma$-algebra, we have $\O \in \Sigma_X$, so in either case we have: 
:$\paren {S_1 \times S_2}^y \in \Sigma_X$
That is: 
:$S_1 \times S_2 \in \FF$
It remains to show that $\FF$ is a $\sigma$-algebra.
Since $S_1 \times S_2 \in \FF$ for $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.
Since $X \in \Sigma_X$ and $Y \in \Sigma_Y$, we obtain: 
:$X \times Y \in \FF$
We show that $\FF$ is closed under countable union.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\FF$.
We have $\paren {E_n}^y \in \Sigma_X$ for each $n \in \N$. 
So, since $\Sigma_X$ is a $\sigma$-algebra, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}^y \in \Sigma_X$
From Union of Horizontal Sections is Horizontal Section of Union, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}^y = \paren {\bigcup_{n \mathop = 1}^\infty E_n}^y$ 
So we have: 
:$\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n}^y \in \Sigma_X$
That is: 
:$\ds \bigcup_{n \mathop = 1}^\infty E_n \in \FF$
We finally show that $\FF$ is closed under complementation.
Let $E \in \FF$.
We then have $E^y \in \Sigma_X$.
Since $\Sigma_X$ is closed under complementation, we have $X \setminus E^y \in \Sigma_X$. 
From Complement of Horizontal Section of Set is Horizontal Section of Complement, we have: 
:$X \setminus E^y = \paren {\paren {X \times Y} \setminus E}^y$
so that: 
:$\paren {\paren {X \times Y} \setminus E}^y \in \Sigma_X$
giving: 
:$\paren {X \times Y} \setminus E \in \FF$
So $\FF$ is a $\sigma$-algebra.
As discussed, we therefore obtain: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$ 
In particular, for any $E \in \Sigma_X \otimes \Sigma_Y$, we have $E \in \FF$.
That is: 
:for any $E \in \Sigma_X \otimes \Sigma_Y$ we have $E^y \in \Sigma_X$
as was the demand.
Category:Horizontal Section of Measurable Set is Measurable
Horizontal Section of Measurable Set is Measurable/Proof 2

== Theorem ==
== Proof ==
Define $f_y : X \to X \times Y$ by: 
:$\map {f_y} x = \tuple {x, y}$ 
for each $x \in X$.
Note that we have $\tuple {x, y} \in E$ $x \in E^y$ from the definition of the horizontal section.
In other words: 
:$\paren {f_y}^{-1} \sqbrk E = E^y$
We now show that $f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.
We use Mapping Measurable iff Measurable on Generator.
By the definition of the product $\sigma$-algebra, $\Sigma_X \otimes \Sigma_Y$ is generated by the rectangles $A \times B$ for $A \in \Sigma_X$ and $B \in \Sigma_Y$. 
From Horizontal Section of Cartesian Product, we have:
:$\ds \paren {A \times B}^y = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$
So for each $A \in \Sigma_X$ and $B \in \Sigma_Y$, we have: 
:$\ds \paren {f_y}^{-1} \sqbrk {A \times B} = \begin{cases}A & y \in B \\ \O & y \not \in B\end{cases}$
From the definition of a $\sigma$-algebra, we have:
:$\O \in \Sigma_X$
So we have: 
:$\ds \paren {f_y}^{-1} \sqbrk {A \times B} \in \Sigma_X$
for any $A \in \Sigma_X$ and $B \in \Sigma_Y$. 
So for any $G \in \set {A \times B : A \in \Sigma_X, \, B \in \Sigma_Y}$ we have: 
:$\paren {f_y}^{-1} \sqbrk G \in \Sigma_X$ 
So by Mapping Measurable iff Measurable on Generator:
:$f_y$ is $\Sigma_X/\paren {\Sigma_X \otimes \Sigma_Y}$-measurable.
So, for any $E \in \Sigma_X \otimes \Sigma_Y$ we have: 
:$\paren {f_y}^{-1} \sqbrk E \in \Sigma_X$
So:
:$E^y \in \Sigma_X$
Category:Horizontal Section of Measurable Set is Measurable
Horizontal Section of Simple Function is Simple Function

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y}$ be the product measurable space of $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$.
Let $f : X \times Y \to \R$ be a simple function.
Let $y \in Y$. 
Then $f^y : X \to \R$ is a simple function, where $f^y$ is the $y$-horizontal section of $f$. 
== Proof ==
Write the standard representation of $f$ as: 
:$\ds f = \sum_{k \mathop = 1}^n a_k \chi_{E_k}$
with:
:$E_1, E_2, \ldots, E_n$ pairwise disjoint $\Sigma_X \otimes \Sigma_Y$-measurable sets
:$a_1, a_2, \ldots, a_n$ real numbers.
We have: 
From Intersection of Horizontal Sections is Horizontal Section of Intersection, we have that: 
:$\paren {E_1}^y, \paren {E_2}^y, \ldots, \paren {E_n}^y$ are pairwise disjoint.
From Horizontal Section of Measurable Set is Measurable, we have that: 
:$\paren {E_1}^y, \paren {E_2}^y, \ldots, \paren {E_n}^y$ are $\Sigma_X$-measurable.
So, we have: 
:$\ds f^y = \sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}^y}$
with:
:$\paren {E_1}^y, \paren {E_2}^y, \ldots, \paren {E_n}^y$ pairwise disjoint $\Sigma_X$-measurable sets
:$a_1, a_2, \ldots, a_n$ real numbers.
So $f^y$ is a simple function.
Category:Horizontal Section of Functions
Category:Simple Functions
Horizontal Section preserves Increasing Sequences of Functions

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\sequence {f_n}_{n \mathop \in \N}$ be an increasing sequence of real-valued functions with $f_i : X \times Y \to \overline \R$ for each $i$. 
Let $y \in Y$. 
Then the sequence $\sequence {\paren {f_n}^y}_{n \mathop \in \N}$ is increasing, where $\paren {f_n}^y$ denotes the $y$-horizontal section of $f_n$. 
== Proof ==
Since $\sequence {f_n}_{n \mathop \in \N}$ is an increasing sequence of real-valued functions, we have: 
:$\map {f_i} {x, y} \le \map {f_j} {x, y}$ for all $i, j$ with $i \le j$. 
for all $\tuple {x, y} \in X \times Y$.
In particular, for fixed $y \in Y$, we have: 
:$\map {f_i} {x, y} \le \map {f_j} {x, y}$ for all $i, j$ with $i \le j$ and all $x \in X$.
From the definition of the $y$-horizontal section, we have: 
:$\map {\paren {f_i}^y} x \le \map {\paren {f_j}^y} x$ for all $i, j$ with $i \le j$ and all $x \in X$.
So: 
:$\sequence {\paren {f_n}^y}_{n \mathop \in \N}$ is an increasing sequence of real-valued functions.
Category:Horizontal Section of Functions