Functions and Sets 
Robert, Susan, and Thomas are the sole contestants in a lottery in which two prizes will be awarded. Three tickets with their names on them are placed in a hat. The person whose name is on the first ticket drawn wins prize #1. That ticket is placed back in. Then the second ticket is drawn. You can win both prizes. Let $A = \{1, 2\}$, $B = \{r, s, t\}$ where $R = \text {robert}$, etc
a.) Explain how each possible assignment of prizes to contestants may be thought of as a function from $A$ to $B$ and why $B^A$ may be thought of as representing the set of all such possible assignments.
d.) In terms of assignments of prizes in the lottery, what does it mean to say that an element of $B^A$ is an injection?

My attempt: 
a.) So it's essentially 
f(1) = r, s, t
f(2) = r, s, t
Because either prize can map to any contestant. But then these aren't functions, are they? Because the same value maps to multiple people (potentially).
And the function A --> B is the set of all possible assignments because it explains how awards can be matched to contestants? 
d.) So an injective function is 1 to 1. If an element of A --> B is an injection, that means each value in A (domain) only maps to one value in the codomain (B), and no two values in A map to the same value in B. That is to say, an element of B^A that is an injection means that a player did not win both of the prizes.

Am I vaguely close?

For a,it is OK for a function to map different inputs to the same output. For functions of the reals, $f(x)=0$ is a fine function-it takes any real as an input and returns $0$.  A function just has to return a single value for any input.  You haven't done the last sentence of the problem, showing how $B^A$ can represent all assignments.
For d, "each value in A (domain) only maps to one value in the codomain (B)" is required of any function, so is not peculiar to injective functions and could be deleted.  The rest of the answer is good.
