We are given the relation $R \subset A \times B$, where $A = \{1\}$ and $B = \{1, 2, 3\}$, and the relation $R = \{(1,1), (1,2), (1,3)\}$.

Let's check the conditions:

1. Is $R$ a function?

A relation $R$ is a function if each element of the domain $A$ is related to exactly one element of the codomain $B$.

• Here, $A = \{1\}$, and the element $1 \in A$ is mapped to multiple elements in $B$: $1$, $2$, and $3$.
• Since the element $1 \in A$ is mapped to more than one element in $B$, $R$ is not a function.

2. Is $R$ injective?

A relation (or function) is injective (one-to-one) if different elements in the domain map to different elements in the codomain. However, for injectivity, we require that the relation is first a function.

• Since $R$ is not a function, we do not check for injectivity. Hence, $R$ is not injective.

3. Is $R$ surjective?

A relation (or function) is surjective (onto) if every element in the codomain $B$ has at least one preimage in the domain $A$.

• Here, the codomain $B = \{1, 2, 3\}$, and every element in $B$ is mapped from the element $1 \in A$ (i.e., $1 \to 1$, $1 \to 2$, and $1 \to 3$).
• Therefore, $R$ is surjective, as every element of $B$ has a preimage in $A$.

Conclusion:

• $R$ is surjective.
• $R$ is not a function.
• $R$ is not injective.

The correct options are:

• $R$ is surjective
• $R$ is not a function

Would you like more details on any part of this explanation?

Here are 5 related questions:

1. What makes a relation a function in set theory?
2. How is injectivity different from surjectivity in functions?
3. Can a relation be surjective without being a function?
4. How does the Cartesian product $A \times B$ influence the nature of a relation?
5. In what cases can a relation be both injective and surjective?

Tip: For a relation to be a function, each input (from the domain) must correspond to exactly one output (in the codomain).