nyuuzyou/wikis huggingface dataset:
Projection from Metric Space Product with P-Product Metric is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d}$ and $M_2 = \struct {A_2, d'}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d_p}$ denote the metric space on $\AA$ where $d_p: \AA \to \R$ is the $p$-product metric on $\AA$:
:$\map {d_p} {x, y} := \paren {\paren {\map d {x_1, y_1} }^p + \paren {\map {d'} {x_2, y_2} }^p}^{1/p}$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to M_2$ denote the first projection and second projection respectively on $\MM$.
Then $\pr_1$ and $\pr_2$ are both ‎continuous on $\MM$.
== Proof ==
We want to show that, for all $a \in \AA$:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map {d_p} {z, a} < \delta \implies \map d {\map {\pr_1} z, \map {\pr_1} a} < \epsilon$
and:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map {d_p} {z, a} < \delta \implies \map {d'} {\map {\pr_2} z, \map {\pr_2} a} < \epsilon$
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $a = \tuple {x_0, y_0} \in \AA$ also be arbitrary.
Let $\delta = \epsilon$.
Let $z = \tuple {x_1, y_1} \in \AA$ such that $\map {d_p} {x, a} < \delta$.
We have:
Then:
and similarly:
Hence:
and:
We have that $a$ and $\epsilon$ are arbitrary.
Hence the result by definition of ‎continuity.
Category:Projections
Category:P-Product Metrics
Category:Continuous Mappings on Metric Spaces
Projection from Metric Space Product with Taxicab Metric is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d}$ denote the metric space on $\AA$ where $d: \AA \to \R$ is the taxicab metric on $\AA$:
:$\map d {x, y} := \map {d_1} {x_1, y_1} + \map {d_2} {x_2, y_2}$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to M_2$ denote the first projection and second projection respectively on $\MM$.
Then $\pr_1$ and $\pr_2$ are both ‎continuous on $\MM$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 15$
Category:Projections
Category:Taxicab Metric
Category:Continuous Mappings on Metric Spaces
Category:Projection from Metric Space Product with Taxicab Metric is Continuous
Projection from Metric Space Product with Taxicab Metric is Continuous/Proof 1

== Theorem ==
== Proof ==
The taxicab metric is an instance of the $p$-product metric:
:$\map {d_p} {x, y} := \paren {\paren {\map {d_1} {x_1, y_1} }^p + \paren {\map {d_2} {x_2, y_2} }^p}^{1/p}$
where $p = 1$.
The result is therefore seen to be an instance of Projection from Metric Space Product with P-Product Metric is Continuous.
Category:Projection from Metric Space Product with Taxicab Metric is Continuous
Projection from Metric Space Product with Taxicab Metric is Continuous/Proof 2

== Theorem ==
== Proof ==
We want to show that, for all $a \in \AA$:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_1} {\map {\pr_1} z, \map {\pr_1} a} < \epsilon$
and:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_2} {\map {\pr_2} z, \map {\pr_2} a} < \epsilon$
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $a = \tuple {x_0, y_0} \in \AA$ also be arbitrary.
Let $\delta = \epsilon$.
Let $z = \tuple {x_1, y_1} \in \AA$ such that $\map d {x, a} < \delta$.
We have:
Hence:
and:
We have that $a$ and $\epsilon$ are arbitrary.
Hence the result by definition of ‎continuity.
Category:Projection from Metric Space Product with Taxicab Metric is Continuous
Projection from Product Category

== Theorem ==
Let $\CC, \DD$ be categories.
Let $\CC \times \DD$ be the product category.
Then the projection functors:
:$\pi_\CC: \CC \times \DD \to \CC$
:$\pi_\DD: \CC \times \DD \to \DD$
are indeed functors.
Moreover, $\pi_\CC, \pi_\DD$ satisfy the following universal property:
:For any category $\EE$ and any functors $F : \EE \to \CC$, $G : \EE \to \DD$, there exists a unique functor $H : \EE \to \CC \times \DD$ such that $F = \pi_\CC H$ and $G = \pi_\DD H$
That is, the following diagram commutes:
:$\xymatrix {
\CC & \CC \times \DD \ar[l]_{\pi_\CC} \ar[r]^{\pi_\DD} & \DD \\
& \EE \ar[lu]^F \ar[u]_{\exists ! H} \ar[ru]_G}$
== Proof ==
Category:Category Theory
Category:Projections
Projection from Product Topology is Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are are continuous.
=== General Result ===
== Proof ==
From Natural Basis of Product Topology of Finite Product, a basis for $\tau$ is:
:$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$
Let $U$ be open in $T_1$.
Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology.
Thus $\pr_1$ is continuous.
The same argument can be applied to $\pr_2$.
== Also see ==
* Projection from Product Topology is Open
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products: Proposition $3.5.3$
Category:Product Topology
Category:Continuous Mappings
Category:Projections
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Continuous/General Result

== Theorem ==
Let $\family {T_i}_{i \mathop \in I} = \family {\struct{S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the Tychonoff topology on $S$.
Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.
Then $\pr_i$ is continuous for all $i \in I$.
== Proof ==
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Continuous/General Result/Proof

== Theorem ==
== Proof ==
By definition of the product topology on $S$:
:$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the Initial Topoplogy:Definition 2:
:$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous.
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Open

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are open.
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Projection from Product Topology is Continuous
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 34 \ \text {(a)}$
Category:Product Topology
Category:Open Mappings
Category:Projections
Category:Projection from Product Topology is Open and Continuous
Category:Projection from Product Topology is Open
Projection from Product Topology is Open/General Result

== Theorem ==
Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the product topology on $S$.
Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.
Then $\pr_i$ is open for all $i \in I$.
== Proof ==
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Open/General Result/Proof

== Theorem ==
== Proof ==
Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$
where:
:$J$ is an arbitrary index set
:$n_j \in \N$
:$i_{k, j} \in I$
:$U_{k, j} \in \tau_{i_{k, j} }$.
For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by:
:$V_{i', k, j} = \begin {cases} U_{k, j} & : i' = i_{k, j} \\ S_{i'} & : i' \ne i_{k, j} \end {cases}$
For all $i \in I$, we have:
As:
:$\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} \in \tau_i$
it follows that $\pr_i$ is open.
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Open/Proof 1

== Theorem ==
== Proof ==
Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$
where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k, j} \in \set {1, 2}$, and $U_{k, j} \in \tau_{i_{k, j} }$.
For all $i \in \set {1, 2}$, define $V_{i, k, j} \in \tau_i$ by:
:$V_{i, k, j} = \begin {cases} U_{k, j} & : i = i_{k, j} \\ S_i & : i \ne i_{k, j} \end {cases}$
By definition of projection:
:$\map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } = \ds \prod_{i \mathop = 1}^2 V_{i, k, j}$
, we consider $\pr_1$.
We have:
As $\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j} \in \tau_1$, it follows that $\pr_1$ is open.
The proof for $\pr_2$ is the same, .
Category:Projection from Product Topology is Open
Projection from Product Topology is Open/Proof 2

== Theorem ==
== Proof ==
We show that $\pi_1$ is open, the proof for $\pi_2$ is identical.
Let $V$ be open in $T$.
We show that $\pi_1 \sqbrk V$ is open in $T_1$.
From Set is Open iff Neighborhood of all its Points, it is enough to show that for each $x \in \pi_1 \sqbrk V$, there exists an open neighborhood $U_1$ of $x$ such that $U_1 \subseteq \pi_1 \sqbrk V$.
Let $x \in \pi_1 \sqbrk V$.
Then there exists $y \in S_2$ such that $\tuple {x, y} \in V$. 
From Natural Basis of Product Topology, there exists an open neighborhood $U_1 \subseteq X$ of $x$ and an open neighborhood $U_2 \subseteq Y$ of $y$ such that $U_1 \times U_2 \subseteq V$. 
Then we have $U_1 \subseteq \pi_1 \sqbrk V$.
Since $x$ was arbitrary, from Set is Open iff Neighborhood of all its Points it follows that $\pi_1 \sqbrk V$ is open.
Category:Projection from Product Topology is Open
Projection from Product Topology is Open and Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are open and continuous.
=== General Result ===
== Proof ==
=== Projection is Continuous ===
=== Projection is Open ===
Category:Product Topology
Category:Continuous Mappings
Category:Open Mappings
Category:Projections
Category:Projection from Product Topology is Open and Continuous
Projection from Product Topology is Open and Continuous/General Result

== Theorem ==
Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the product topology on $S$.
Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.
Then $\pr_i$ is open and continuous for all $i \in I$.
== Proof ==
=== Projection is Continuous ===
=== Projection is Open ===
Category:Projection from Product Topology is Open and Continuous
Projection from Topological Product is Identification Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then $\pr_1$ and $\pr_2$ are identification mappings.
== Proof ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 40$
Category:Identification Topology
Category:Product Topology
Projection in Plane between Lines passing through Origin is Linear Operator

== Theorem ==
Let $M$ and $N$ be distinct straight lines through the plane through the origin.
Let $\pr_{M, N}$ be the projection on $M$ along $N$.
Then $\pr_{M, N}$ is a linear operator.
== Proof ==
Let the angle between $M$ and the $x$-axis be $\theta$.
Let the angle between $N$ and the $x$-axis be $\phi$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then from Equations defining Projection in Plane:
:$\map {\pr_{M, N} } P = \begin {cases} 
\tuple {0, y - x \tan \phi} & : \theta = \dfrac \pi 2 \\
\tuple {x, x \tan \theta} & : \phi = \dfrac \pi 2 \\
\tuple {\dfrac {x \tan \phi} {\tan \phi - \tan \theta} - \dfrac y {\tan \phi - \tan \theta}, \dfrac {y \tan \theta} {\tan \theta - \tan \phi} - \dfrac {x \tan \theta \tan \phi} {\tan \theta - \tan \phi} } & : \text {otherwise} 
\end {cases}$
This demonstrates that $\map {\pr_{M, N} } P$ can be expressed as an ordered tuple of $4$ real numbers.
The result follows from Linear Operator on the Plane.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Linear Operators
Category:Geometric Projections
Projection in Plane on X-Axis along Y-Axis

== Theorem ==
Let $\pr_{X, Y}$ denote the projection on the $x$-axis along the $y$-axis:
:$\forall P \in \R^2: \map {\pr_{X, Y} } P =$ the intersection of the $x$-axis with the line through $P$ parallel to the $y$-axis.
Let $P = \tuple {\lambda_1, \lambda_2}$ be an arbitrary point in $\R^2$.
Then:
:$\map {\pr_{X, Y} } {\lambda_1, \lambda_2} = \tuple {\lambda_1, 0}$
== Proof ==
This is an instance of a projection on $M$ along $N$ where $N$ coincides with the $y$-axis.
Hence it is one of the special cases of Equations defining Projection in Plane: Cartesian:
:$\map {\pr_{M, N} } P = \tuple {\lambda_1, \lambda_1 \tan \theta}$
where $\theta$ is the angle between $M$ and the $x$-axis.
In this case, $M$ is actually coincident with the $x$-axis.
Hence $\theta = 0$ and so from Tangent of Zero:
:$\map {\pr_{X, Y} } P = \tuple {\lambda_1, 0}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Geometric Projections
Projection in Plane on Y-Axis along X-Axis

== Theorem ==
Let $\pr_{Y, X}$ denote the projection on the $y$-axis along the $x$-axis:
:$\forall P \in \R^2: \map {\pr_{Y, X} } P =$ the intersection of the $y$-axis with the line through $P$ parallel to the $x$-axis.
Let $P = \tuple {\lambda_1, \lambda_2}$ be an arbitrary point in $\R^2$.
Then:
:$\map {\pr_{Y, X} } {\lambda_1, \lambda_2} = \tuple {0, \lambda_2}$
== Proof ==
This is an instance of a projection on $M$ along $N$ where $M$ coincides with the $y$-axis.
Hence it is one of the special cases of Equations defining Projection in Plane: Cartesian:
:$\map {\pr_{M, N} } P = \tuple {0, \lambda_2 - \lambda_1 \tan \phi}$
where $\phi$ is the angle between $N$ and the $x$-axis.
In this case, $N$ is actually coincident with the $x$-axis.
Hence $\phi = 0$ and so from Tangent of Zero:
:$\map {\pr_{X, Y} } P = \tuple {0, \lambda_2}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Geometric Projections
Projection is Epimorphism

== Theorem ==
Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Then:
:$\pr_1$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_1, \circ_1}$
:$\pr_2$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_2, \circ_2}$
where $\pr_1$ and $\pr_2$ are the first and second projection respectively of $\struct {\SS, \circ}$.
=== General Result ===
== Proof ==
From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections.
We now need to show they are homomorphisms.
Let $s, t \in \struct {\SS, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$.
Then:
Thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.3$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.3$
Category:Projection is Epimorphism
Category:Epimorphisms (Abstract Algebra)
Category:Projections
Projection is Epimorphism/General Result

== Theorem ==
Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_k, \circ_k}, \ldots, \struct {S_n, \circ_n}$.
Then:
:for each $j \in \closedint 1 n$, $\pr_j$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_j, \circ_j}$
where $\pr_j: \struct {\SS, \circ} \to \struct {S_j, \circ_j}$ is the $j$th projection from $\struct {\SS, \circ}$ to $\struct {S_j, \circ_j}$.
== Proof ==
From Projection is Surjection, $\pr_j$ is a surjection for all $j$.
We now need to show it is a homomorphism.
Let $\mathbf s, \mathbf t \in \struct {\SS, \circ}$ where $\mathbf s = \tuple {s_1, s_2, \ldots, s_j, \ldots, s_n}$ and $\mathbf t = \tuple {t_1, t_2, \ldots, t_j, \ldots, t_n}$.
Then:
and thus the morphism property is demonstrated.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.12$
* : $\S 3.1$: Direct sums
Category:Projection is Epimorphism
Projection is Injection iff Factor is Singleton

== Theorem ==
Let $S_1, S_2, \ldots, S_n$ be non-empty sets.
Let $\ds S = \prod_{i \mathop = 1}^n S_i$ be the cartesian product of $S_1, S_2, \ldots, S_n$.
Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.
Then $\pr_j$ is an injection $S_k$ is a singleton for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
=== Family of Sets ===
== Proof ==
=== Sufficient Condition ===
Suppose $S_k = \set {s_k}$ for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
Let $\map {\pr_j} x = \map {\pr_j} y = z$ for $x, y \in S$.
Then by definition of $j$th projection, $x, y \in S$ are given by:
:$x = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$
:$y = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$
and so $x = y$.
Thus $\pr_j$ is an injection by definition.
=== Necessary Condition ===
Let $\pr_j$ be an injection.
Thus:
:$\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$
Let $x, y \in S$ be such that:
:$x = \tuple {x_1, x_2, \dotsc, x_{j - 1}, z, x_{j + 1}, \dotsc, x_n}$
:$y = \tuple {y_1, y_2, \dotsc, y_{j - 1}, z, y_{j + 1}, \dotsc, y_n}$
Then:
:$\map {\pr_j} x = \map {\pr_j} y = z$
For $x = y$ it is necessary that:
:$x_1 = y_1, x_2 = y_2, \dotsc, x_{j - 1} = y_{j - 1}, x_{j + 1} = y_{j + 1}, \dotsc, x_n = y_n$
That is:
:$\forall k \in \set {1, 2, \dotsc, n}, k \ne j: x_k = y_k$
That is, that $S_k$ is a singleton for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
== Sources ==
* : $\S 8$: Functions
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $5$
Category:Projections
Category:Injections
Category:Singletons
Category:Projection is Injection iff Factor is Singleton
Projection is Injection iff Factor is Singleton/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is an arbitrary index set.
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.
Then $\pr_j$ is an injection $S_i$ is a singleton for all $i \in I \setminus \set j$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
Category:Projection is Injection iff Factor is Singleton
Projection is Injection iff Factor is Singleton/Family of Sets/Necessary Condition

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is an arbitrary index set.
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.
Let $\pr_j$ be an injection.
Then $S_i$ is a singleton for all $i \in I \setminus \set j$.
== Proof ==
Let $\pr_j$ be an injection.
Then:
:$\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$
We have that $\family {S_i}_{i \mathop \in I}$ is a non-empty family of non-empty sets
Hence, by the axiom of choice (formulation $2$), $S$ is non-empty.
Let $z \in S$.
By the definition of Cartesian product $S$:
:$\forall i \in I: \map z i \in S_i$.
Let $x_k \in S_k$ for some $k \in I \setminus \set j$.
Let $z' \in S$ be defined by:
:$\map {z'} i = \begin{cases}
\map z i & : i \in I \setminus \set k \\
x_k & : i = k 
\end{cases}$
Then:
Thus:
:$z' = z$
In particular:
:$x_k = \map {z'} k = \map z k = z_k$
It follows that:
:$S_k = \set {z_k}$
Since $k$ was arbitrary:
:$\forall k \in I \setminus \set j: S_k = \set {z_k}$
The result follows.
Category:Projection is Injection iff Factor is Singleton
Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is an arbitrary index set.
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.
Let $S_i$ be a singleton for all $i \in I \setminus \set j$.
Then $\pr_j$ is an injection.
== Proof ==
Let $S_i = \set {s_i}$ for all $i \in I \setminus \set {j}$.
Let $\map {\pr_j} x = \map {\pr_j} y = s_j$ for $x, y \in S$.
By definition of $j$th projection:
:$\map x j = \map {\pr_j} x = s_j$
:$\map y j = \map {\pr_j} y = s_j$
and so $\map x j = \map y j$.
By the definition of Cartesian product, for all $i \in I \setminus \set j$:
:$\map x i, \map y i \in S_i = \set {s_i}$
and so $\map x i = \map y i$ for all $i \in I \setminus \set j$.
Thus $x = y$.
Thus $\pr_j$ is an injection by definition.
Category:Projection is Injection iff Factor is Singleton
Projection is Surjection

== Theorem ==
Let $S$ and $T$ be non-empty sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Let $\pr_1: S \times T \to S$ and $\pr_2: S \times T \to T$ be the first projection and second projection respectively on $S \times T$.
Then $\pr_1$ and $\pr_2$ are both surjections.
=== General Version ===
=== Family of Sets ===
== Proof ==
Let $S$ and $T$ be sets such that neither is empty.
Let $\pr_1: S \times T \to S$ be the first projection on $S \times T$.
Then by definition of first projection:
:$\forall x \in S: \exists \tuple {x, t} \in S \times T: \map {\pr_1} {x, t} = x$
Similarly, let $\pr_2: S \times T \to T$ be the second projection on $S \times T$.
Then by definition of second projection:
:$\forall x \in T: \exists \tuple {s, x} \in S \times T: \map {\pr_2} {s, x} = x$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{R}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.5$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $5$
Category:Surjections
Category:Projections