avewright/proofwiki-math huggingface dataset:
Theorem: Volume of Paraboloid

The volume of paraboloid is half the volume of its circumscribing cylinder.

Proof:
{{MissingLinks}}
420pxright
Consider a cylinder of radius $r$ and height $h$, circumscribing a paraboloid $y = h \paren {\dfrac x r}^2$ whose apex is at the center of the bottom base of the cylinder and whose base is the top base of the cylinder.
Also consider the paraboloid $y = h - h \paren {\dfrac x r}^2$, with equal dimensions but with its apex and base flipped.
For every height $0 \le y \le h$, the disk-shaped cross-sectional area $\pi \paren {r \sqrt {1 - \dfrac y h} }^2$ of the flipped paraboloid is equal to the ring-shaped cross-sectional area $\pi r^2 - \pi r^2 \paren {\sqrt {\dfrac y h} }^2$ of the cylinder part outside the inscribed paraboloid.
Indeed:
{{begin-eqn}}
{{eqn | l = \pi \paren {r \sqrt {1 - \dfrac y h} }^2
      | r = \pi r^2 \paren {\sqrt {1 - \dfrac y h} }^2
      | c = 
}}
{{eqn | r = \pi r^2 \paren {1 - \dfrac y h}
      | c = 
}}
{{eqn | r = \pi r^2 - \pi r^2 \dfrac y h
      | c = 
}}
{{eqn | r = \pi r^2 - \pi r^2 \paren {\sqrt {\dfrac y h} }^2
      | c = 
}}
{{end-eqn}}
{{mistake|discrepancy}}
Therefore, the volume of the flipped paraboloid is equal to the volume of the cylinder part outside the inscribed paraboloid.
In other words, the volume of the  paraboloid is $\dfrac \pi 2 r^2 h$, half the volume of its circumscribing cylinder.
{{ProofWanted}}
Category:Paraboloids
Category:Volume Formulas
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Theorem: Quotient Mapping equals Surjective Identification Mapping

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a mapping.
Then $f$ is a quotient mapping, {{Iff}}:
:$f$ is surjective, and $\tau_2$ is the identification topology on $S_2$ with respect to $f$ and $T_1$.

Proof:
=== Sufficient condition ===
Let $f$ be surjective, and $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $T_1$.
Identification Mapping is Continuous shows that $f$ is continuous.
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of identification topology, it follows that $U$ is open in $T_2$.
Hence $f$ fulfills all the requirements to be a quotient mapping.
{{qed|lemma}}
Theorem: Quotient Topology is Topology

Let $\struct {S, \tau}$ be a topological space.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Let $\tau_\RR$ be the quotient topology on $S / \RR$ by $q_\RR$.
Then $\tau_\RR$ is a topology on $S$.

Proof:
By definition of quotient topology, $\tau_\RR$ is the identification topology on $S / \RR$ with respect to $q_\RR$ and $\struct {S, \tau}$.
Identification Topology is Topology shows that $\tau_\RR$ is a topology on $S / \RR$.
{{qed}}
Theorem: Quotient Mapping Induces Homeomorphism between Quotient Space and Image

Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a quotient mapping.
Let $\RR_f \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Let $q_{\RR_f}: S_1 \to S_1 / \RR_f$ be the quotient mapping induced by $\RR_f$.
Let $\tau_{\RR_f}$ be the quotient topology on $S_1 / \RR_f$ by $q_{\RR_f}$.
Then there exists a uniquely defined homeomorphism $\tilde f: S_1 / \RR_f \to S_2$ such that $f = \tilde f \circ q_{\RR_f}$.
The homeomorphism between the quotient space $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$ is defined by:
:$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$

Proof:
Quotient Mapping equals Surjective Identification Mapping shows that $\tau_2$ is the identification topology on $S_2$ with respect to $f$ and $\struct { S_1, \tau_1}$.
Identification Topology equals Quotient Topology on Induced Equivalence shows that $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$ are homeomorphic.
The proof of Identification Topology equals Quotient Topology on Induced Equivalence shows that a homeomorphism $\tilde f: S_1 / \RR_f \to S_2$ is defined by:
:$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$
As we have $\map { q_{\RR_f} } s = \eqclass s { \RR_f}$, it follows that $\tilde f \circ q_{\RR_f} = f$.
To show uniqueness, let $g: S_1 / \RR_f \to S_2$ be a mapping such that $g \circ q_{\RR_f} = f$.
For all $t \in S_2$, we can find $s \in S_1$ such that $\map f s = t$, as $f$ is surjective by definition of quotient mapping.
Then:
:$\map { g \circ q_{\RR_f} } s = \map f s = \map { \tilde f \circ q_{\RR_f} } s$
and as $\map {q_{\RR_f} } s$ is uniquely defined, we must have $\map g {\eqclass s { \RR_f} } = \map {\tilde f}{\eqclass s { \RR_f} }$.
As this is true for all $\map g {\eqclass s { \RR_f} } \in S_2$, it follows that $g = \tilde f$.
{{qed}}
Category:Quotient Mappings (Topology)
Category:Quotient Spaces (Topology)
acpmyilecdoxme411631ymhi66fw5yj
Theorem: Continuous Closed Surjective Mapping is Quotient Mapping

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a continuous closed surjective mapping.
Then $f$ is a quotient mapping.

Proof:
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
For $X \subseteq S$, let $\relcomp S X$ denote the relative complement of $X$ is $S$.
By definition of closed set, $\relcomp {S_1} {f^{-1} \sqbrk U}$ is closed in $T_1$.
By definition of closed mapping, $f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} }$ is closed in $T_2$.
By definition of closed set, $\relcomp {S_2} {f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} } }$ is open in $T_2$.
Then:
{{begin-eqn}}
{{eqn | l = \relcomp {S_2} {f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} } }
      | r = \relcomp {S_2} {f \sqbrk {f^{-1} \sqbrk {\relcomp {S_2} U} } }
      | c = Complement of Preimage equals Preimage of Complement
}}
{{eqn | r = \relcomp {S_2} {\relcomp {S_2} U}
      | c = {{Corollary|Image of Preimage under Mapping}}
}}
{{eqn | r = U
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
By definition of quotient mapping, it follows that $f$ is a quotient mapping.
{{qed}}
Theorem: Upper Bound for Lowest Common Multiple

Let $a, b \in \Z$ be integers such that $a b \ne 0$.
Then:
:$\lcm \set {a, b} \le \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$

Proof:
By Product of GCD and LCM:
:$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
By Existence of Greatest Common Divisor $\gcd \set {a, b}$ exists.
By definition of GCD, $\gcd \set {a, b} \in \Z_{>0}$.
Hence the result.
{{qed}}
Theorem: Dilation of Convex Set in Vector Space is Convex

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a vector space over $\Bbb F$. 
Let $C \subseteq X$ be a convex subset of $X$.
Let $\alpha \in \Bbb F$. 
Then $\alpha C$ is convex.

Proof:
Consider first the case $\alpha = 0$.
We then have $\alpha C = \set { {\mathbf 0}_X}$.
This is convex by Singleton is Convex Set.
Now consider the case $\alpha \ne 0$. 
Let $u, v \in \alpha C$ and $t \in \closedint 0 1$. 
Then there exists $x, y \in C$ such that $u = \alpha x$ and $v = \alpha y$.
Since $C$ is convex, we have: 
:$t x + \paren {1 - t} y \in C$ 
Then, we have: 
:$\alpha t x + \alpha \paren {1 - t} y \in \alpha C$ 
That is: 
:$t u + \paren {1 - t} v = t \paren {\alpha x} + \paren {1 - t} \paren {\alpha y} \in \alpha C$
Since $u, v \in \alpha C$ and $t \in \closedint 0 1$ were arbitrary, $\alpha C$ is convex.
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Convex Sets (Vector Spaces)
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Theorem: Continuous Open Surjective Mapping is Quotient Mapping

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a continuous open surjective mapping.
Then $f$ is a quotient mapping.

Proof:
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of open mapping:
:$f \sqbrk {f^{-1} \sqbrk U}$ is open in $T_2$
From {{Corollary|Image of Preimage under Mapping}}:
:$f \sqbrk {f^{-1} \sqbrk U} = U$
It follows that $U$ is open in $T_2$.
By definition of quotient mapping, it follows that $f$ is a quotient mapping.
{{qed}}
Theorem: Composite of Quotient Mappings in Topology is Quotient Mapping

Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be quotient mappings.
Then $g \circ f : S_1 \to S_3$ is a quotient mapping.

Proof:
Composite of Surjections is Surjection shows that $g \circ f$ is surjective.
Composite of Continuous Mappings is Continuous shows that $g \circ f$ is continuous.
Let $U \subseteq S_3$ such that $\paren {g \circ f}^{-1} \sqbrk U$ is open in $T_1$.
By definition of quotient mapping:
:$f \sqbrk {\paren {g \circ f}^{-1} \sqbrk U}$ is open in $T_2$.
By definition of quotient mapping:
:$g \circ f \sqbrk {\paren {g \circ f}^{-1} \sqbrk U}$ is open in $T_3$.
From {{Corollary|Image of Preimage under Mapping}}:
:$g \circ f \sqbrk {\paren {g \circ f}^{-1} \sqbrk U} = U$
It follows that $U$ is open in $T_3$.
Therefore, $g \circ f : S_1 \to S_3$ is a continuous surjective mapping that fulfills the following condition:
:For all $U \subseteq S_3$, if $\paren {g \circ f}^{-1} \sqbrk U$ is open in $T_1$, then $U$ is open in $T_3$.
It follows that $g \circ f$ is a quotient mapping by definition.
{{qed}}
Theorem: Quotient Mapping and Continuous Mapping Induces Continuous Mapping

Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.
Let $p: S_1 \to S_2$ be a quotient mapping.
Let $g: S_2 \to S_3$ be a mapping such that for all $s_1, s_2 \in S_1$ with $\map p {s_1} = \map p {s_2}$, we have $\map g {s_1} = \map g {s_2}$.
Then $g$ induces a mapping $f: S_2 \to S_3$ such that $f \circ p = g$.
The induced mapping $f$ is a continuous mapping {{iff}} $g$ is a continuous mapping.

Proof:
For all $t \in S_2$ we can find $s \in S_1$ with $\map p s = t$, as $p$ is surjective.
Define $\map f t := \map g s$.
For all $s' \in S_1$ with $\map p {s'} = t$, we have $\map g s = \map g {s'}$ {{hypothesis}}.
Hence $f$ is well-defined.
It follows that $\map g s = \map {f \circ p} s$.
Suppose $f$ is continuous.
As $p$ is continuous by definition of quotient mapping, Composite of Continuous Mappings is Continuous shows that $g = f \circ p$ is continuous.
Suppose $g$ is continuous.
Let $V \subseteq S_3$ be open in $T_3$.
As $g$ is continuous:
:$g^{-1} \sqbrk V$ is open in $T_1$.
As $g = f \circ p$:
:$g^{-1} \sqbrk V = p^{-1} \sqbrk {f^{-1} \sqbrk V}$
Hence $p^{-1} \sqbrk {f^{-1} \sqbrk V}$ is open in $T_1$. 
As $p$ is a quotient mapping:
:$f^{-1} \sqbrk V$ is open in $T_2$.
As $f^{-1} \sqbrk V$ is open in $T_2$:
:$f$ is continuous.
{{qed}}
Theorem: Set of Integers with GCD of 1 are not necessarily Pairwise Coprime

Let $S$ be a set of integers such that $S$ has more than $2$ elements:
:$S = \set {s_1, s_2, \ldots, s_n}$
Let:
:$\map \gcd S = 1$
where $\gcd$ denotes the GCD of $S$.
Then it is not necessarily the case that there exist a pair of elements of $S$ which are themselves pairwise coprime:
:$\exists i, j \in \set {1, 2, \ldots, n}: \gcd \set {s_i, s_j} = 1$

Proof:
Proof by Counterexample
Let $S = \set {6, 10, 15}$.
We have:
{{begin-eqn}}
{{eqn | l = \gcd \set {6, 10}
      | r = 2
      | c = 
}}
{{eqn | l = \gcd \set {6, 15}
      | r = 3
      | c = 
}}
{{eqn | l = \gcd \set {10, 15}
      | r = 5
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Quotient Mapping and Continuous Mapping Induces Continuous Mapping/Corollary

The induced mapping $f$ is a quotient mapping, {{iff}} $g$ is a quotient mapping.

Proof:
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $f$ is well-defined.
Suppose $f$ is a quotient mapping.
Composite of Quotient Mappings in Topology is Quotient Mapping shows that $g = f \circ p$ is a quotient mapping.
Suppose $g$ is a quotient mapping.
It follows that $g$ is surjective.
That is, for all $t \in S_3$ there exists $s \in S_1$ such that $\map g s = t$.
As $t = \map g s = \map { f \circ p } s$, it follows that $f$ is surjective.
By definition of quotient mapping, $g$ is continuous.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $f$ is continuous.
Let $U \subseteq S_3$ such that $f^{-1} \sqbrk U$ is open in $T_2$.
By definition of quotient mapping, $p$ is continuous.
By definition of continuity, it follows that $p^{-1} \sqbrk {f^{-1} \sqbrk U}$ is open in $T_1$.
As $f \circ p = g$, it follows that $p^{-1} \sqbrk {f^{-1} \sqbrk U} = g^{-1} \sqbrk U$.
By definition of quotient mapping, $U$ is open in $T_3$.
By definition of quotient mapping, $f$ is a quotient mapping.
{{qed}}
Theorem: Injective Quotient Mapping Equals Homeomorphism

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a mapping.
Then:
:$f$ is an injective quotient mapping
{{iff}}:
:$f$ is a homeomorphism.

Proof:
=== Sufficient condition ===
Suppose $f$ is an injective quotient mapping.
By definition of quotient mapping, $f$ is surjective.
Mapping is Injection and Surjection iff Inverse is Mapping shows that $f$ has an inverse $f^{-1}$.
To show continuity of $f^{-1}$, let $U \subseteq S_1$ be open in $T_1$.
As $f$ is bijective, it follows that $f^{-1} \sqbrk {f \sqbrk U} = U$.
By definition of quotient mapping, it follows that $f \sqbrk U$ is open in $T_2$.
As $f$ is bijective, it follows that the preimage of $U$ under $f^{-1}$ is equal to $f \sqbrk U$.
It follows that $f^{-1}$ is continuous.
By definition of homeomorphism, it follows that $f$ is a homeomorphism.
{{qed|lemma}}
Theorem: Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step

Let the following hold:
:For all $j \le k$, a tree of order $j$ is of size $j - 1$.
Then this holds:
:A tree of order $k + 1$  is of size $k$.

Proof:
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Take any node $v$ of $T_{k + 1}$ of degree $1$.
Such a node exists from Finite Tree has Leaf Nodes.
Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.
By Connected Subgraph of Tree is Tree, $T_k$ is itself a tree.
The order of $T_k$ is $k$, and it has one less edge than $T_{k + 1}$ by definition.
We have {{hypothesis}} that $T_k$ has $k - 1$ edges.
So $T_{k + 1}$ must have $k$ edges.
Theorem: Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step

Let the following hold:
:For all $j \le k$, a tree of order $j$ is of size $j - 1$.
Then this holds:
:A tree of order $k + 1$  is of size $k$.

Proof:
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Remove any edge $e$ of $T_{k + 1}$.
By definition of tree $T_{k + 1}$ has no circuits.
Therefore from Condition for Edge to be Bridge it follows that $e$ must be a bridge.
So removing $e$ disconnects $T_{k + 1}$ into two trees $T_1$ and $T_2$, with $k_1$ and $k_2$ nodes, where $k_1 + k_2 = k + 1$.
Then, {{hypothesis}}, $T_1$ and $T_2$ have $k_1 - 1$ and $k_2 - 1$ edges.
Putting the edge $e$ back again, it can be seen that $T_{k + 1}$ has $\paren {k_1 - 1} + \paren {k_2 - 1} + 1 = k$ edges.
Therefore a tree of order $k + 1$  is of size $k$.
Theorem: Semi-Inner Product with Zero Vector

Let $\struct {V, \innerprod \cdot \cdot}$ be a semi-inner product space.
Let $\mathbf 0_V$ be the zero vector of $V$.
Then for all $v \in V$:
:$\innerprod {\mathbf 0_V} v = \innerprod v {\mathbf 0_V} = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \innerprod {\mathbf 0_V} v
      | r = \innerprod {0 \cdot \mathbf 0_V} v
}}
{{eqn | r = 0 \cdot \innerprod {\mathbf 0_V} v
      | c = Semi-Inner Product Axioms: Sesquilinearity
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \innerprod v {\mathbf 0_V}
      | r = \overline {\innerprod {\mathbf 0_V} v}
      | c = Semi-Inner Product Axioms: Conjugate Symmetry
}}
{{eqn | r = \overline 0
      | c = From above
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
Theorem: Continuous Surjection Induces Continuous Bijection from Quotient Space

Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $g: S_1 \to S_2$ be a $\tuple {\tau_1, \tau_2}$-continuous surjection.
Let $\RR_g \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $g$:
:$\tuple {s_1, s_2} \in \RR_g \iff \map g {s_1} = \map g {s_2}$
Let $q_{\RR_g}: S_1 \to S_1 / \RR_g$ be the canonical surjection induced by the equivalence relation $\RR_g$.
Let $\tau_{\RR_g}$ be the quotient topology on the quotient space $S_1 / \RR_g$ by $q_{\RR_g}$.
Then $g$ induces a $\tuple {\tau_{\RR_g}, \tau_2}$-continuous bijection $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.
<nowiki>
$\qquad \begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{ q_{\RR_g} }
     \ar[rd]_*{g}
&
 S_1 / \RR_g \ar@{-->}[d]^*{f}
\\ &
 S_2
}\end{xy}$
</nowiki>

Proof:
By definition of quotient topology, $q_{\RR_g}$ is a surjective identification mapping.
Quotient Mapping equals Surjective Identification Mapping shows that $q_{\RR_g}$ is a quotient mapping.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $g$ induces a continuous mapping $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.
As $g$ is surjective, it follows that $f$ is surjective.
Let $\eqclass {s_1} {\RR_g}, \eqclass {s_2} {\RR_g} \in S_1 / \RR_g$ such that $\map f {\eqclass {s_1} {\RR_g} } = \map f {\eqclass {s_2}{ \RR_g} }$.
As $f \circ q_{\RR_g} = g$, it follows that $\map g {s_1} = \map g {s_2}$.
By definition of equivalence relation induced by $g$, it follows that $\eqclass {s_1} {\RR_g} = \eqclass {s_2} {\RR_g}$.
By definition of injection, it follows that $f$ is injective.
As $f$ is injective and surjective, it follows that $f$ is bijective.
{{qed}}
Theorem: Vector Space of Sequences with Finite Support is Vector Space

Let $\struct {K, +_K, \circ_K}$ be a division ring.
Let $\struct{V, +, \cdot}$ be the vector space of sequences with finite support in $K$.
Then $V$ is a vector space over $K$.

Proof:
Consider $V$ as a subset of the vector space of all mappings from $\N$ to $K$.
Let us apply the One-Step Vector Subspace Test.
Thus, let $\sequence{a_n}, \sequence{b_n} \in V$ be sequences over $K$ with finite support, and let $\lambda \in K$.
Then we need to show that $\sequence{ a_n + \lambda b_n }$ has finite support.
It is immediate that if:
:$a_n = b_n = 0$
then:
:$a_n + \lambda b_n = 0$ 
Therefore:
:$\set{ n \in \N: a_n + \lambda b_n \ne 0} \subseteq \set{ n \in \N: a_n \ne 0 } \cup \set{ n \in \N: b_n \ne 0 }$
By Union of Finite Sets is Finite, it follows that $\sequence{ a_n + \lambda b_n }$ has finite support.
The result follows by the One-Step Vector Subspace Test.
{{qed}}
Category:Examples of Vector Spaces
a4hqccu3df5ihsy8cfcvi9sm677g90x
Theorem: Expression for bilinear function

Let $f$ be a real function of two independent variables, $f \in \R \times \R \to \R$.
Then:
:$\map f {x, y}$ is a linear function of $x$ when $y$ is equal to a real constant
:$\map f {x, y}$ is a linear function of $y$ when $x$ is equal to a real constant
{{iff}} $f$ has the form:
:$\exists a, b, c, d \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$

Proof:
=== Sufficient Condition ===
Let:
:$\map f {x, y}$ be a linear function of $x$ when $y$ is equal to a real constant
:$\map f {x, y}$ be a linear function of $y$ when $x$ is equal to a real constant
We need to show that:
:$\exists a, b, c, d \in \R: \forall {x, y} \in \R: \map f {x, y} = a x y + b x + c y + d$
Theorem: Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 1

The induced mapping $f$ is a homeomorphism, {{iff}} $g$ is a quotient mapping.

Proof:
Continuous Surjection Induces Continuous Bijection from Quotient Space shows that $g$ is well-defined.
Let $f$ be a homeomorphism.
Injective Quotient Mapping Equals Homeomorphism shows that $f$ is a quotient mapping.
Composite of Quotient Mappings in Topology is Quotient Mapping shows that $g = f \circ q_{\RR_g}$ is a quotient mapping.
{{qed|lemma}}
Let $g$ be a quotient mapping.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping/Corollary shows that $f$ is a quotient mapping.
Injective Quotient Mapping Equals Homeomorphism shows that $f$ is a homeomorphism.
{{qed}}
Theorem: Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 2

If $S_2$ is a Hausdorff space, then $S_1 / \RR_g$ is a Hausdorff space.

Proof:
Continuous Surjection Induces Continuous Bijection from Quotient Space shows there exists a continuous bijection $f: S_1 / \RR_g \to S_2$.
Let $\eqclass {s_1}{\RR_g}, \eqclass {s_2}{\RR_g} \in S_1 / \RR_g$ such that $\eqclass {s_1}{\RR_g} \ne \eqclass {s_2}{\RR_g}$.
As $f$ is injective, $\map f { \eqclass {s_1}{\RR_g} } \ne \map f { \eqclass {s_2}{\RR_g} }$.
As $S_2$ is a Hausdorff space, there exists disjoint open sets $U_1, U_2 \in \tau_2$ such that $\map f { \eqclass {s_1}{\RR_g} } \in U_1$ and $\map f { \eqclass {s_2}{\RR_g} } \in U_2$.
It follows that $f^{-1} \sqbrk {U_1}$ and $f^{-1} \sqbrk {U_2}$ are disjoint.
As $f$ is continuous, we have $f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2} \in \tau_{\RR_g}$.
By definition of Hausdorff space, $S_1 / \RR_g$ is a Hausdorff space.
{{qed}}
Theorem: Linear Transformation is Fredholm Operator iff Pseudoinverse exists

Let $U, V$ be vector spaces.
Let $T: U \to V$ be a linear transformation.
Then $T$ is a Fredholm operator {{iff}} $T$ has a pseudoinverse.

Proof:
Recall the definitions:
$S$ and $T$ are '''pseudoinverse''' to each other {{iff}}:
:$T \circ S - I_U$ is degenerate
and:
:$S \circ T - I_V$ is degenerate.
$T$ is a '''Fredholm operator''' {{iff}}:
:$(1): \quad \map \ker T$ is finite-dimensional
:$(2): \quad$ the quotient space $V / \Img T$ is finite-dimensional.
Theorem: Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval

Let $\struct {X, \tau_X }$ be a Hausdorff space.
Let $\gamma : \closedint 0 1 \to X$ be a simple loop.
Let $\sim$ be an equivalence relation on $\closedint 0 1$ defined by:
{{begin-eqn}}
{{eqn | o =
      | r = \forall t_1 \in \openint 0 1 , t_2 \in \closedint 0 1 :
      | rr = t_1 \sim t_2 \iff t_2 = t_1
}}
{{eqn | o =
      | r = \forall t_1 \in \set {0, 1} , t_2 \in \closedint 0 1 :
      | rr = t_1 \sim t_2 \iff t_2 \in \set {0,1}
}}
{{end-eqn}}
Let $q : \closedint 0 1 \to \closedint 0 1 / \sim$ be the canonical surjection induced by $\sim$.
Let $\tau_q$ be the quotient topology on the quotient space $\closedint 0 1 / \sim$ induced by $q$.
Let $\tau_\gamma$ be the subspace topology on $\Img \gamma$ induced by $\tau_X$, where $\Img \gamma$ denotes the image of $\gamma$.
Then $\struct {\Img \gamma, \tau_\gamma}$ is homeomorphic to $\struct {\closedint 0 1 / \sim , \tau_q}$.

Proof:
Define $\tilde \gamma: \closedint 0 1 \to \Img \gamma$ as the restriction of $\gamma$ to $\closedint 0 1 \times \Img \gamma$.
Restriction of Mapping to Image is Surjection shows that $\tilde \gamma$ is surjective.
Subspace of Hausdorff Space is Hausdorff shows that $\struct {\Img \gamma, \tau_\gamma}$ is a Hausdorff space.
Closed Real Interval is Compact Space shows that $\closedint 0 1$ is compact.
Continuous Mapping from Compact Space to Hausdorff Space is Closed Mapping shows that $\tilde \gamma$ is a closed mapping.
Continuous Closed Surjective Mapping is Quotient Mapping shows that $\tilde \gamma$ is a quotient mapping.
By definition of simple loop, the equivalence relation $\sim$ is the equivalence induced by $\tilde \gamma$.
Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 1 shows that there exists a homeomorphism $f: 
\closedint 0 1 / \sim \to \Img \gamma$.
::<nowiki>$\begin{xy} \xymatrix@L+2mu@+1em{
 \closedint 0 1 \ar[r]^*{q}
     \ar[rd]_*{\tilde \gamma}
&
 \closedint 0 1 / \sim \ar@{-->}[d]^*{f}
\\ &
 \Img \gamma
}\end{xy}$</nowiki>
{{qed}}
Category:Quotient Spaces (Topology)
Category:Loops (Topology)
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