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  1. This is neither injective nor surjective. It is not injective because more than one element from the domain has 3 as its image. It is not surjective because there are elements of the codomain (1, 2, 4, and 5) that are not images of anything from the domain.

  2. This is a bijection. Every element in the codomain is the image of exactly one element of the domain.

  3. This is a bijection. Note that we can write this function in two line notation as \(f = \twoline{1 \amp 2 \amp 3 \amp 4 \amp 5}{5 \amp 4 \amp 3 \amp 2 \amp 1}\text{.}\)

  4. In two line notation, this function is \(f = \twoline{1 \amp 2 \amp 3 \amp 4 \amp 5}{1 \amp 1 \amp 2 \amp 2 \amp 3}\text{.}\) From this we can quickly see it is neither injective (for example, 1 is the image of both 1 and 2) nor surjective (for example, 4 is not the image of anything).

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