Define the function f: R -> (0, 1], where f(x) = e^(-x^2). Is this map surjective only, injective only, bijective, or none of these options?

The function f is neither surjective, injective, nor bijective.

Here's why:

1. Surjectivity: A function is surjective (or onto) if for every element in the codomain, there is at least one element in the domain that maps to it. In this case, the codomain is the interval (0, 1]. However, not every element in the codomain has a preimage in the domain. For example, there is no real number x such that f(x) = 0. Therefore, the function is not surjective.

2. Injectivity: A function is injective (or one-to-one) if every element in the codomain is mapped to by exactly one element in the domain. In this case, the function f is not injective. For example, consider the real numbers x = 1 and x = -1. We have f(1) = e^(-1) and f(-1) = e^(-1). So, the same element in the codomain is the image of two different elements in the domain. Therefore, the function is not injective.

3. Bijectivity: A function is bijective if it is both surjective and injective. Since the function f is neither surjective nor injective, it is not bijective.

So, the function f is neither surjective, injective, nor bijective.