Discrete Mathematics - Injective Function Justification

Question

Good Evening Stack,

I would just like to know whether or not I'm on the right track with this question as I'm having second thoughts about another possible interpretation.

Question

Define a function f : P({0, 1}) × P({1, 2}) → {0, 1, 2} by f((A, B)) =| A ∩ B |

Is the function f injective? Justify your answer.

Working So Far

Let A = P({0,1}) and B = P({1,2})

There are four possible pairs of F((A,B)). With these being...

F((A1,B1)) = F((0,1)) , F((A2,B2)) = F((0,2)) , F((A3,B3)) = F((1,1)) and F((A4,B4)) = F((1,2)).

Thus, equating the following cardinalities to be...

| A1 ∩ B1 | = 0 . | A2 ∩ B2 | = 0 . | A3 ∩ B3 | = 1 . | A4 ∩ B4 | = 0

I'm unsure as what to do with all of this information, at first I intepretted it as elements could not be mapped unless they had a cardinality >1, but know I believe that the function itself is being mapped to the answer provided by the cardinality, So in that case.

Each pair maps to either the cardinality of 0 or 1. Thus making the statement not injective but surjective,

Would really appreciate a hand, p.s sorry if any of this is hard to read it is my first year of uni so my notation is probrably unformal.

Thank You! Liam.

Answer 1

Note that $\mathcal{P}(\{0, 1\}) = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\}$ and that $\mathcal{P}(\{1, 2\}) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$.

If $f$ is injective, then $f(A, B) = f(C, D) \implies A = C \wedge B = D$. However, $f(\{0, 1\}, \{1, 2\}) = \{0, 1\} \cap \{1, 2\} = \{1\} = \{1\} \cap \{1\} = f(\{1\}, \{1\})$, so $f$ is not injective.

Answer 2

If a function is injective, then the size of its domain is less than or equal to the size of its range. Here the domain of the function has $16$ elements, but the range has $3$.