nyuuzyou/wikis huggingface dataset:
Pullback Commutes with Generating Partition

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA \subseteq \Sigma$ be a finite sub-$\sigma$-algebra.
Let $T: X \to X$ be a measurable mapping.
Let $n \in \N$.
Then:
:$T^{-n} \map \xi \AA = \map \xi {T^{-n} \AA}$
where:
:$\map \xi \cdot$ denotes the generated finite partition
:$\map {T^{-n}} \AA$ denotes the pullback partition of $\AA$ by $T^n$
:$T^{-n} \map \xi \AA$ denotes the pullback partition of $\map \xi \AA$ by $T^n$
== Proof ==
== Sources ==
* $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Pullback Connection is Connection on Smooth Manifold

== Theorem ==
Let $M$ and $\tilde M$ be smooth manifolds with or without boundaries.
Let $\phi : M \to \tilde M$ be a diffeomorphism.
Let $T \tilde M$ and $TM$ be tangent bundles of $\tilde M$ and $M$ respectively.
Let $\tilde \nabla$ be the connection in $T \tilde M$.
Let $\phi^* \tilde \nabla$ be the pullback connection.
Then $\phi^* \tilde \nabla$ is the connection in $TM$.
== Proof ==
== Sources ==
* : $\S 4$: Connections. Pullback Connections
Category:Pullbacks
Category:Connections
Pullback Functor is Functor

== Theorem ==
Let $\mathbf C$ be a metacategory having all pullbacks.
Let $f: C \to D$ be a morphism of $\mathbf C$.
Let $\mathbf C \mathbin / C$ and $\mathbf C \mathbin / D$ be the slice categories over $C$ and $D$, respectively.
Let $f^* : \mathbf C \mathbin / D \to \mathbf C \mathbin / C$ be the pullback functor defined by $f$.
Then $f^*$ is a functor.
== Proof ==
Let $\alpha: A \to D$ be an object of $\mathbf C \mathbin / D$.
Then the identity morphism $\operatorname{id}_\alpha: \alpha \to \alpha$ is by definition $\operatorname{id}_A: A \to A$.
Thus $f^* \operatorname{id}_\alpha$ is by definition the unique morphism fitting:
::$\begin{xy}\xymatrix@+1em{
 A'
 \ar[rr]^*{f_\alpha}
 \ar[dd]_*{f^* \alpha}
 \ar@{-->}[rd]^*{f^* \mathrm{id}_\alpha}
& &
 A
 \ar[rd]^*{\mathrm{id}_\alpha}
 \ar[dd]^(.4)*{\alpha}
\\ &
 A'
 \ar[ld]^*{f^* \alpha}
 \ar[rr] |{\hole} ^(.3)*{f_\alpha}
& &
 A
 \ar[ld]^*{\alpha}
\\
 C
 \ar[rr]_*{f}
& &
 D
}\end{xy}$
It is apparent that $\operatorname{id}_{A'}$ satisfies this condition.
Since by definition, $\operatorname{id}_{f^* \alpha} = \operatorname{id}_{A'}$, this shows that:
:$f^* \operatorname{id}_\alpha = \operatorname{id}_{f^* \alpha}$
Now let $\gamma_1: \alpha_2 \to \alpha_1$ and $\gamma_2: \alpha_3 \to \alpha_2$ be morphisms of $\mathbf C \mathbin / D$.
Then $f^* \left({\gamma_1 \circ \gamma_2}\right)$ is the unique morphism fitting the dashed arrow in:
::$\begin{xy}\xymatrix@+1em{
 A_3'
 \ar[rr]^*{f_{\alpha_3}}
 \ar[dd]_*{f^* \alpha_3}
 \ar@{-->}[rd]
& &
 A_3
 \ar[rd]^*{\gamma_1 \circ \gamma_2}
 \ar[dd]^(.4)*{\alpha_3}
\\ &
 A_1'
 \ar[ld]^*{f^* \alpha_1}
 \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}}
& &
 A_1
 \ar[ld]^*{\alpha_1}
\\
 C
 \ar[rr]_*{f}
& &
 D
}\end{xy}$
Unfolding the composition $\gamma_1 \circ \gamma_2$ yields the expanded diagram:
::$\begin{xy}\xymatrix@+1em{
 A_3'
 \ar[rr]^*{f_{\alpha_3}}
 \ar[ddd]_*{f^* \alpha_3}
 \ar[rd]^*{f^* \gamma_2}
 \ar@{-->}@/_/[rdd]
& &
 A_3
 \ar[rd]^*{\gamma_2}
 \ar[ddd]^*{\alpha_3}
\\ &
 A_2'
 \ar[d]^*{f^* \gamma_1}
 \ar[rr] |{\hole} ^(.3)*{f_{\alpha_2}}
& &
 A_2
 \ar[d]^*{\gamma_1}
\\ &
 A_1'
 \ar[ld]^*{f^* \alpha_1}
 \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}}
& &
 A_1
 \ar[ld]^*{\alpha_1}
\\
 C
 \ar[rr]_*{f}
& &
 D
}\end{xy}$
in which $f^* \left({\gamma_1 \circ \gamma_2}\right)$ again fits the dashed arrow.
But this means, in particular, that we must have:
:$f^* \left({\gamma_1 \circ \gamma_2}\right) = f^* \gamma_1 \circ f^* \gamma_2$
Therefore, $f^*$ is a functor.
== Sources ==
* : $\S 5.3$: Proposition $5.10$
Category:Functors
Category:Pullbacks
Pullback Lemma

== Theorem ==
Let $\mathbf C$ be a metacategory.
Suppose the following is a commutative diagram in $\mathbf C$:
::$\begin{xy}\xymatrix@+0.5em{
 F
 \ar[d]_*+{h}
 \ar[r]^*+{f'}
&
 E
 \ar[d]^*+{h'}
 \ar[r]^*+{g'}
&
 D
 \ar[d]^*+{h}
\\
 A
 \ar[r]_*+{f}
&
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
Suppose furthermore that the right square is a pullback.
Then the left square is a pullback the outer rectangle is.
== Proof ==
=== Necessary Condition ===
Suppose the left square is a pullback.
To verify the outer rectangle is a pullback, suppose we are given the following commutative diagram:
::$\begin{xy}\xymatrix@+0.5em{
P
\ar@/^1em/[rrrd]^*+{p_1}
\ar@/_1em/[ddr]_*+{p_2}
\\ & & &
D
\ar[d]^*+{h}
\\ &
A
\ar[r]_*+{f}
&
B
\ar[r]_*+{g}
&
C
}\end{xy}$
The right square being a pullback implies there is a unique $v: P \to E$ , making the following commute:
::$\begin{xy}\xymatrix@+0.5em{
P
\ar@/^1em/[rrrd]^*+{p_1}
\ar@/_1em/[ddr]_*+{p_2}
\ar@{-->}@/^/[rrd]_*+{v}
\\ & &
E
\ar[d]^*+{h'}
\ar[r]^*+{g'}
&
D
\ar[d]^*+{h}
\\ &
A
\ar[r]_*+{f}
&
B
\ar[r]_*+{g}
&
C
}\end{xy}$
Since the left square is also a pullback, we subsequently find a unique $u: P \to F$ , making the following commute:
::$\begin{xy}\xymatrix@+0.5em{
 P
 \ar@/^1em/[rrrd]^*+{p_1}
 \ar@/_1em/[ddr]_*+{p_2}
 \ar@/^/[rrd]_*+{v}
 \ar@{-->}[rd]_*+{u}
\\ &
 F
 \ar[d]_*+{h}
 \ar[r]^*+{f'}
&
 E
 \ar[d]^*+{h'}
 \ar[r]^*+{g'}
&
 D
 \ar[d]^*+{h}
\\ &
 A
 \ar[r]_*+{f}
&
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
In conclusion, the outer rectangle is a pullback.
=== Sufficient Condition ===
Suppose the outer rectangle is a pullback.
To verify the left square is a pullback, suppose we are given the following commutative diagram:
::$\begin{xy}\xymatrix@+0.5em{
 P
 \ar@/^/[rrd]^*+{p_1}
 \ar@/_1em/[ddr]_*+{p_2}
\\ & &
 E
 \ar[d]^*+{h'}
\\ &
 A
 \ar[r]_*+{f}
&
 B
}\end{xy}$
Evidently, we can amend this diagram as follows:
::$\begin{xy}\xymatrix@+0.5em{
 P
 \ar@/^/[rrd]_*{p_1}
 \ar@/^1em/[rrrd]^*+{g' \circ p_1}
 \ar@/_1em/[ddr]_*+{p_2}
\\ & &
 E
 \ar[d]^*+{h'}
 \ar[r]^*+{g'}
&
 D
 \ar[d]^*+{h}
\\ &
 A
 \ar[r]_*+{f}
&
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
Because the outer rectangle is a pullback, we thus obtain a unique $u: P \to F$, making the following commute:
::$\begin{xy}\xymatrix@+0.5em{
 P
 \ar@/^1em/[rrrd]^*+{g' \circ p_1}
 \ar@/_1em/[ddr]_*+{p_2}
 \ar@{-->}[rd]_*+{u}
\\ &
 F
 \ar[d]_*+{h}
 \ar[r]^*+{f'}
&
 E
 \ar[d]^*+{h'}
 \ar[r]^*+{g'}
&
 D
 \ar[d]^*+{h}
\\ &
 A
 \ar[r]_*+{f}
&
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
If we now can prove that $f' \circ u = p_1$, then we're done.
To this end, remark that they both complete the following commutative diagram:
::$\begin{xy}\xymatrix@+0.5em{
 P
 \ar@{-->}[rrd]
 \ar@/^1em/[rrrd]^*+{g' \circ p_1}
 \ar@/_1em/[ddr]_*+{p_2}
\\ & &
 E
 \ar[d]^*+{h'}
 \ar[r]^*+{g'}
&
 D
 \ar[d]^*+{h}
\\ &
 A
 \ar[r]_*+{f}
&
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
Since the right square is a pullback, there is only one such morphism.
Therefore, we conclude $f' \circ u = p_1$.
Hence the left square is also a pullback.
== Sources ==
* : $\S 5.3$: Lemma $5.8$
Category:Named Theorems
Category:Pullbacks
Category:Category Theory
Pullback as Equalizer

== Theorem ==
Let $\mathbf C$ be a metacategory with binary products.
Suppose that the following diagram commutes (possibly except for the square):
$\quad\quad \begin{xy}\xymatrix@+1em{
P
\ar@/^/[rrd]^*+{p_1}
\ar@/_/[ddr]_*+{p_2}
\ar[rd]^*+{e}
\\
&
A \times B
\ar[r]_*+{\pi_1}
\ar[d]^*+{\pi_2}
&
A
\ar[d]^*+{f}
\\
&
B
\ar[r]_*+{g}
&
C
}\end{xy}$
where $A \times B$ is the binary product of $A$ and $B$, and $\pi_1, \pi_2$ are the associated projections.
Then the border of the diagram, i.e.:
$\quad\quad \begin{xy}\xymatrix{
P
\ar[r]^*+{p_1}
\ar[d]_*+{p_2}
&
A
\ar[d]^*+{f}
\\
B
\ar[r]_*+{g}
&
C
}\end{xy}$
is a pullback of $f$ and $g$ $e: P \to A \times B$ is the equalizer of $f \circ \pi_1$ and $g \circ \pi_2$.
=== Corollary ===
== Proof ==
== Sources ==
* : $\S 5.2$: Proposition $5.5$
Category:Pullbacks
Pullback as Equalizer/Corollary

== Corollary to Pullback as Equalizer ==
Let $\mathbf C$ be a metacategory.
Suppose $\mathbf C$ has all binary products and all equalizers.
Then $\mathbf C$ has all pullbacks.
== Proof ==
Let $f: A \to C$ and $g: B \to C$ be morphisms of $\mathbf C$ with common codomain.
By assumption on $\mathbf C$, there exists a binary product $A \times B$ with projections $\pi_1: A \times B \to A$ and $\pi_2: A \times B \to B$.
Again by assumption on $\mathbf C$, there exists an equalizer $e: P \to A \times B$ of $f \circ \pi_1$ and $g \circ \pi_2$.
From Pullback as Equalizer, the pullback of $f$ and $g$ is given by:
::$\begin{xy}\xymatrix@+1em{
 P
 \ar[r]^*+{p_1}
 \ar[d]_*+{p_2}
&
 A
 \ar[d]^*+{f}
\\
 B
 \ar[r]_*+{g}
&
 C
}\end{xy}$
where $p_1 = \pi_1 \circ e$ and $p_2 = \pi_2 \circ e$.
== Sources ==
* : $\S 5.2$: Corollary $5.6$
Category:Pullbacks
Pullback as Limit

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f_1: A \to C$ and $f_2: B \to C$ be morphisms of $\mathbf C$.
Let their pullback:
::$\begin{xy}\xymatrix@+1em@L+3px{
 P
 \ar[r]^*+{p_2}
 \ar[d]_*+{p_1}
&
 A
 \ar[d]^*+{f_1}
\\
 B
 \ar[r]_*+{f_2}
&
 C
}\end{xy}$
exist in $\mathbf C$.
Then $\struct {P, p_1, p_2}$ is the limit of the diagram $D: \mathbf J \to \mathbf C$ defined by:
::$\begin{xy}\xymatrix@+1em@L+3px{
 \save[]+<0em,-2em>*{\mathbf{J}:} \restore
& &
 \cdot
 \ar[d]
\\ &
 \cdot
 \ar[r]
&
 \cdot
}\end{xy}$
::$\begin{xy}\xymatrix@+1em@L+3px{
 \save[]+<0em,-2em>*{D:} \restore
& &
 A
 \ar[d]^*+{f_1}
\\ &
 B
 \ar[r]_*+{f_2}
&
 C
}\end{xy}$
== Proof ==
== Sources ==
* : $\S 5.4$: Example $5.19$
Category:Category Theory
Category:Pullbacks
Pullback of Commutative Triangle

== Theorem ==
Let $\mathbf C$ be a metacategory.
Suppose that the following is a commutative diagram in $\mathbf C$:
::$\begin{xy}\xymatrix@+1em@L+2px{
 A'
 \ar[rr]^*{h_\alpha}
 \ar[dd]_*{\alpha'}
& &
 A
 \ar[rd]^*{\gamma}
 \ar[dd]^(.4)*{\alpha}
\\ &
 B'
 \ar[ld]^*{\beta'}
 \ar[rr] |{\hole} ^(.3)*{h_\beta}
& &
 B
 \ar[ld]^*{\beta}
\\
 C'
 \ar[rr]_*{h}
& &
 C
}\end{xy}$
and that the two squares in it are pullback diagrams.
Then there is a unique morphism $\gamma': A' \to B'$ making the following commute:
::$\begin{xy}\xymatrix@+1em@L+2px{
 A'
 \ar[rr]^*{h_\alpha}
 \ar[dd]_*{\alpha'}
 \ar@{-->}[rd]^*{\gamma'}
& &
 A
 \ar[rd]^*{\gamma}
 \ar[dd]^(.4)*{\alpha}
\\ &
 B'
 \ar[ld]^*{\beta'}
 \ar[rr] |{\hole} ^(.3)*{h_\beta}
& &
 B
 \ar[ld]^*{\beta}
\\
 C'
 \ar[rr]_*{h}
& &
 C
}\end{xy}$
Furthermore, $\gamma'$ makes the following a pullback:
::$\begin{xy}\xymatrix@+.5em@L+2px{
 A'
 \ar[r]^*{h_\alpha}
 \ar[d]_*{\gamma'}
&
 A
 \ar[d]^*{\gamma}
\\
 B'
 \ar[r]_*{h_\beta}
&
 B
}\end{xy}$
== Proof ==
The first diagram above can be distorted into:
::$\begin{xy}\xymatrix@+.5em@L+2px{
 A'
 \ar@/^/[rrd]^*{\gamma \circ h_\alpha}
 \ar@/_/[ddr]_*{\alpha'}
\\ &
 B'
 \ar[r]^*{h_\beta}
 \ar[d]^*{\beta'}
&
 B
 \ar[d]^*{\beta}
\\ &
 C'
 \ar[r]_*{h}
&
 C
}\end{xy}$
and since the square is a pullback, there is a unique $\gamma': A' \to B'$ as desired.
That:
::$\begin{xy}\xymatrix@+.5em@L+2px{
 A'
 \ar[r]^*{h_\alpha}
 \ar[d]_*{\gamma'}
&
 A
 \ar[d]^*{\gamma}
\\
 B'
 \ar[r]_*{h_\beta}
&
 B
}\end{xy}$
is a pullback follows immediately from the Pullback Lemma.
== Sources ==
* : $\S 5.3$: Corollary $5.9$
Category:Pullbacks
Pullback of Quotient Group Isomorphism/Examples

== Examples of Pullbacks of Quotient Group Isomorphisms ==
=== Quotient Groups of Order 2 Subgroups ===
=== Alternating Subgroups of Symmetric Groups ===
=== Cyclic Group $C_6$ with Alternating Group $A_4$ ===
Category:Examples of Pullbacks of Quotient Group Isomorphisms
Pullback of Quotient Group Isomorphism/Examples/Alternating Subgroups of Symmetric Groups

== Example of Pullback of Quotient Group Isomorphism ==
Let $S_m$ and $S_n$ be symmetric groups on $m$ and $n$ letters respectively.
Let $A_m$ and $A_n$ be the alternating groups on $m$ and $n$ letters respectively.
Let $\theta: S_m / A_m \to S_n / A_n$ be an isomorphism.
The pullback of $S_m$ and $S_n$ by $\theta$ is a subset of $S_m \times S_n$ of the form:
:$S_m \times^\theta S_n = \set {\tuple {\rho, \sigma}: \map \sgn \rho = \map \sgn \sigma}$
where $\map \sgn \rho$ denotes the sign of $\rho$.
== Proof ==
From Alternating Group is Normal Subgroup of Symmetric Group, $A_m$ is normal in $S_m$.
Similarly with $A_n$.
From Order of Alternating Group it follows that:
:$\index {S_m} {A_m} = 2$
and similarly for $A_n$.
Hence the result from Pullback of Quotient Group Isomorphism: Subgroups of Index 2.
== Sources ==
* : Chapter $13$: Direct products: Example $13.12$
Category:Examples of Pullbacks of Quotient Group Isomorphisms
Pullback of Quotient Group Isomorphism/Examples/C6 with A4

== Example of Pullback of Quotient Group Isomorphism ==
Let $G = C_6$ be the cyclic group of order $6$:
:$G = \gen x$
Let $H = A_4$ be the alternating group on $4$ letters.
Let $e_G$ and $e_H$ denote the identity elements of $G$ and $H$ respectively.
Let $N$ be the subgroup of $G$:
:$N = \gen {x^3}$
which from Subgroup of Abelian Group is Normal is normal.
Let $K$ be the subgroup of $H$:
:$\set {e_H, \tuple {1 2} \tuple {3 4}, \tuple {1 3} \tuple {2 4}, \tuple {1 4} \tuple {2 3} }$
which from Klein Four-Group is Normal in A4 is normal.
Let $\theta: G / N \to H / K$ be the mapping defined as:
:$\forall M \in G / N: \map \theta M = \begin{cases} K & : M = N \\ \tuple {1 2 3} K & : M = x N \\ \tuple {1 3 2} K & : M = x^2 N \end{cases}$
The pullback of $G$ and $H$ by $\theta$ is:
:$G \times^\theta H = \set {\tuple {e_G, e_H}, \tuple {e_G, \tuple {1 2} \tuple {3 4} }, \tuple {e_G, \tuple {1 3} \tuple {2 4} }, \tuple {e_G, \tuple {1 4} \tuple {2 3} }, \\
\tuple {x^3, e_H}, \tuple {x^3, \tuple {1 2} \tuple {3 4} }, \tuple {x^3, \tuple {1 3} \tuple {2 4} }, \tuple {x^3, \tuple {1 4} \tuple {2 3} }, \\
\tuple {x, \tuple {1 2 3} }, \tuple {x, \tuple {1 3 4} }, \tuple {x, \tuple {2 4 3} }, \tuple {x, \tuple {1 4 2} }, \\
\tuple {x^4, \tuple {1 2 3} }, \tuple {x^4, \tuple {1 3 4} }, \tuple {x^4, \tuple {2 4 3} }, \tuple {x^4, \tuple {1 4 2} }, \\
\tuple {x^2, \tuple {1 3 2} }, \tuple {x^2, \tuple {1 4 3} }, \tuple {x^2, \tuple {2 3 4} }, \tuple {x^2, \tuple {1 2 4} }, \\
\tuple {x^5, \tuple {1 3 2} }, \tuple {x^5, \tuple {1 4 3} }, \tuple {x^5, \tuple {2 3 4} }, \tuple {x^5, \tuple {1 2 4} } }$
== Proof ==
== Sources ==
* : Chapter $13$: Direct products: Exercise $5$
Category:Examples of Pullbacks of Quotient Group Isomorphisms
Pullback of Quotient Group Isomorphism/Examples/Subgroups of Index 2

== Example of Pullback of Quotient Group Isomorphism ==
Let $G$ and $H$ be groups.
Let $N$ and $K$ be normal subgroups of $G$ and $H$ respectively such that:
:their quotient groups $G / N$ and $H / K$ are isomorphic
:their indices are $2$:
::$\index G N = \index H K = 2$
Let $\theta: G / N \to H / K$ be an isomorphism.
The pullback of $G$ and $H$ by $\theta$ is a subset of $G \times H$ of the form:
:$G \times^\theta H = \set {\tuple {g, h}: \paren {g \in N, h \in K} \text { or } \paren {g \notin N, h \notin K} }$
== Proof ==
As $\index G N = \index H K = 2$, it follows that:
:$\order {G / N} = \order {H / K} = 2$
and they are the cyclic group of order $2$.
Let:
: $x \in G: x \notin N$
: $y \in H: y \notin K$
Then:
:$G / N = \gen {x N}$
:$H / K = \gen {y K}$
and we have:
Let $\tuple {g, h} \in G \times^\theta H$.
By definition:
:$G \times^\theta H = \set {\tuple {g, h}: \map \theta {g N} = h K}$
Let $g \in N$.
Then $g N = N$ and so:
: $\map \theta {g N} = h K = K$
So $g \in N \implies h \in K$.
Let $g \notin N$.
Then $g N = x N$ and so:
: $\map \theta {g N} = h K = x K$
So $g \notin N \implies h \notin K$.
Hence the result.
== Sources ==
* : Chapter $13$: Direct products: Example $13.12$
Category:Examples of Pullbacks of Quotient Group Isomorphisms
Pullback of Quotient Group Isomorphism is Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity element is $e_G$.
Let $\struct {H, *}$ be a group whose identity element is $e_H$.
Let $N \lhd G, K \lhd H$ be normal subgroups of $G$ and $H$ respectively.
Let:
:$G / N \cong H / K$
where:
:$G / N$ denotes the quotient of $G$ by $N$
:$\cong$ denotes group isomorphism.
Let $\theta: G / N \to H / K$ be such a group isomorphism.
Let $G \times^\theta H$ be the pullback of $G$ and $H$ via $\theta$.
Then $G \times^\theta H$ is a subgroup of $G \times H$.
== Proof ==
This result is proved by an application of the Two-Step Subgroup Test:
=== Condition $(1)$ ===
From the definition of pullback:
:$\tuple {e_G, e_H} \in G \times^\theta H$
:
:$\map \theta {e_G \circ N} = e_H * K$
By Coset by Identity, $e_G \circ N, e_H * K$ are the identities of $G / N$ and $H / K$
From Group Homomorphism Preserves Identity:
:$\map \theta {e_G \circ N} = e_H * K$
So $\tuple {e_G, e_H} \in G \times^\theta H$
Thus $G \times^\theta H$ is non-empty.
=== Condition $(2)$ ===
Let $\tuple {g, h}$ and $\tuple {g', h'}$ be elements of $G \times^\theta H$.
It follows by definition of $\theta$ that:
:$\map \theta {g \circ N} = h * K$
and:
:$\map \theta {g' \circ N} = h' * K$
By the morphism property:
:$\map \theta {g \circ N} * \map \theta {g' \circ N} = \map \theta {g \circ N \circ g' \circ N} = \map \theta {\paren {g \circ g'} \circ N}$
Hence:
Thus:
:$\tuple {g \circ g', h * h'} \in G \times^\theta H$
Hence $G \times^\theta H$ is closed under the operation.
=== Condition $(3)$ ===
Let $\tuple {g, h} \in G \times^\theta H$.
Then:
Then:
Thus $G \times^\theta H$ is closed under inverses.
Therefore by the Two-Step Subgroup Test:
:$G \times^\theta H \le G \times H$
== Sources ==
* : Chapter $13$: Direct products: Definition $13.11$
Category:Pullbacks of Quotient Group Isomorphisms
Category:Quotient Groups
Category:Group Isomorphisms
Category:Subgroups
Pullback of Riemannian Metric by Smooth Mapping is Riemannian Metric iff Mapping is Immersion

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a Riemannian manifold with or without boundary.
Let $M$ be a smooth manifold with or without boundary.
Let $F : M \to \tilde M$ be a smooth mapping.
Let $F^* \tilde g$ be the pullback of $\tilde g$ by $F$.
Let $g = F^* \tilde g$ be a smooth $2$-tensor field.
Then $g$ is a Riemannian metric $F$ is an immersion.
== Proof ==
== Sources ==
* : $\S 2$. Riemannian Metrics. Definitions
Category:Riemannian Geometry
Pullback of Subset Inclusion

== Theorem ==
Denote with $\mathbf{Set}$ the category of sets.
Let $A, B$ be sets, and let $f: A \to B$ be a mapping.
Let $V \subseteq B$ be a subset of $B$.
Denote with $i: V \to B$ the inclusion mapping.
Let $f^{-1} \left({V}\right) \subseteq A$ be the preimage of $V$ under $f$.
Denote with $j: f^{-1} \left({V}\right) \to A$ the inclusion mapping.
Denote with $\bar f = f \restriction_{f^{-1} \left({V}\right)}$ the restriction of $f$ to $f^{-1} \left({V}\right)$.
Then:
$\quad\quad \begin{xy}\xymatrix{
f^{-1} \left({V}\right)
\ar[r]^*+{\bar f}
\ar[d]_*+{j}
&
V
\ar[d]^*+{i}
\\
A
\ar[r]_*+{f}
&
B
}\end{xy}$
is a pullback diagram in $\mathbf{Set}$.
== Proof ==
From the definition of pullback, given a commutative diagram:
$\quad\quad \begin{xy}\xymatrix{
Q
\ar[r]^*+{q_1}
\ar[d]_*+{q_2}
&
V
\ar[d]^*+{i}
\\
A
\ar[r]_*+{f}
&
B
}\end{xy}$
we need to find a mapping $u: Q \to \map {f^{-1}} V$ such that $\bar f \circ u = q_1$ and $j \circ u = q_2$.
Let $x \in Q$; then $\map {q_1} x \in V \subseteq B$.
Note that the commutativity of the diagram implies that $\map {q_1} x = \map f {\map {q_2} x}$ since $i$ is an inclusion.
Hence also $\map f {\map {q_2} x} \in V$.
It follows by definition of preimage that $\map {q_2} x \in \map {f^{-1}} V$.
Thus define $u: Q \to \map {f^{-1}} V$ by restriction of $q_2$:
:$\map u x := \map {q_2 \restriction_{Q \times \map {f^{-1}} V}} x$
Then we have:
:$\map {\paren {j \circ u}} x = \map u x = \map {q_2} x$
since $j$ is an inclusion mapping.
Hence $q_2 = j \circ u$ by Equality of Mappings.
Furthermore:
:$\map {\bar f \circ u} x = \map {\bar f \circ q_2} x = \map f {\map {q_2} x} = \map {q_1} x$
since $\bar f$ and $u$ are restrictions of $f$ and $q_2$, respectively.
Hence $\bar f \circ u = q_1$ by Equality of Mappings.
Observe that any $u': Q \to \map {f^{-1}} V$ must satisfy:
:$j \circ u' = q_2 = j \circ u$
By Inclusion Mapping is Injection, $j$ is an injection.
From Injection iff Monomorphism in Category of Sets, we conclude $j$ is a monomorphism, and so $u = u'$.
This establishes that $u: Q \to \map {f^{-1}} V$ is unique.
Therefore, we have established that:
$\quad\quad \begin{xy}\xymatrix{
\map {f^{-1}} V
\ar[r]^*+{\bar f}
\ar[d]_*+{j}
&
V
\ar[d]^*+{i}
\\
A
\ar[r]_*+{f}
&
B
}\end{xy}$
is a pullback diagram in $\mathbf{Set}$.
== Sources ==
* : $\S 5.2$: Example $5.7$
Category:Pullbacks
Category:Category of Sets
Category:Subsets
Category:Inclusion Mappings
Pumping Lemma for Regular Languages

== Theorem ==
Let $\LL_3$ be the set of regular languages.
Then the following holds:
$\forall L \in \LL_3: \exists n_0 \in \N_0: \forall z \in L: \card z > n_0 \implies \exists u, v, w$ such that:
:$z = u \cdot v \cdot w$
:$\card v > 0$
:$\card {u v} < n_0$
:$\forall i \in \N_0: u \cdot v^i \cdot w \in L$
== Proof ==
=== For finite languages ===
For any finite regular language $L_{fin}$, the proof is simple.
Let $s_{maxlen} \in L_{fin}$.
Thus:
:$\forall s \in L_{fin}: \card s \le \card {s_{maxlen} }$
Now choose $n_0 > \card {s_{maxlen} }$.
The implication now trivially holds because the premise:
:$\paren {\card z > n_0}$
is false.
=== For infinite languages ===
For any one infinite regular language $L_{inf}$, the following holds:
:$(1): \quad$ There exists a finite automaton:
::::$F = \struct {\Sigma, Q, q_0, A, \delta}$
:::such that:
::::$\map \LL F = L_{inf}$.
See Equivalence of Finite Automata and Regular Languages for a demonstration of this.
:$(2): \quad$ There exists an infinite number of words $s \in L_{inf}$ with:
::::$\card s \ge \card Q$
Let $s \in L_{inf}$ be one of these words.
Let $n_0 = \card Q + 1$.
Consider all prefixes of this $s$.
We have:
:$\card s > \card Q$
Thus from the Pigeonhole Principle, for at least two of them $F$ needs to end in the same state.
Let the first two indices in $s$ for which this is the case be $i$ and $j$.
The difference between $prefix_i$ and $prefix_j$ ($v$ in this formulation of the theorem) is a "loop" in $F$.
This loop can be traversed any number of times (including 0) and $F$ will still be in the same state immediately afterwards.
Hence:
:$\forall i \N_0: u v^i w \in L_{inf}$
Furthermore:
:$\card {u v} = j - 1 < n_0$
And since $i \ne j$ it follows that:
:$\card v > 0$
 
=== General case ===
Since it could be shown that the theorem holds for both all finite and all infinite regular languages, it can be stated that it holds for all regular languages.
== Also see ==
* Formal Language is not necessarily Regular, which follows from this result.
Category:Computer Science
Category:Named Theorems
Pursuit Curve of Boat in River

== Theorem ==
Consider a straight river $R$ whose parallel banks are aligned with the $y$-axis and the line $x = c$ of a cartesian plane.
Let the current of $R$ have a constant and uniform speed $a$ in the negative $y$ direction.
Let a boat $B$ be launched from the point $\tuple {c, 0}$ and headed directly towards the origin with speed $b$ relative to the water.
The path of $B$ is defined by the equation:
:$c^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$
where $k = \dfrac a b$.
== Proof ==
The components of the velocity of $b$ are:
:$\dfrac {\d x} {\d t} = - b \cos \theta$
:$\dfrac {\d y} {\d t} = - a + b \sin \theta$
Hence:
$(1)$ is a homogeneous differential equation.
Let $z = \dfrac y x$.
Then by Solution to Homogeneous Differential Equation:
Substituting back for $z$, putting $k = \dfrac a b$, and rearranging:
:$C^k \paren {y + \sqrt {x^2 + x^2} } = x^{k + 1}$
== Sources ==
* : $\S 2.12$: The Hanging Chain. Pursuit Curves: Example $(4)$
Category:Pursuit Curves