# Mapping from Set to Ordinate of Cartesian Product is Injection

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## Contents

## Theorem

Let $S$ and $T$ be sets such that $T \ne \O$.

Let $S \times T$ denote their cartesian product.

Let $t \in T$ be given.

Let $j_t \subseteq S \times \paren {S \times T}$ be the mapping from $S$ to $S \times T$ defined as:

- $\forall s \in \S: \map {j_t} s = \tuple {s, t}$

Then $j_t$ is an injection.

## Proof

It has been shown in Correspondence between Set and Ordinate of Cartesian Product is Mapping that $j_t$ is a mapping.

Now it is to be shown that $j_t$ is injective, that is:

- $\forall s_1, s_2 \in S: \map {j_t} {s_1} = \map {j_t} {s_2} \implies s_1 = s_2$

We have that:

\(\displaystyle \map {j_t} {s_1}\) | \(=\) | \(\displaystyle \map {j_t} {s_2}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \tuple {s_1, t}\) | \(=\) | \(\displaystyle \tuple {s_2, t}\) | Definition of $j_t$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1\) | \(=\) | \(\displaystyle s_2\) | Definition of Ordered Pair |

Hence the result.

$\blacksquare$

## Also see

- Definition:Canonical Injection (Abstract Algebra) for an instance of this construct in the context of algebraic structures

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $6$