To determine if a function is injective, we need to check if each input value maps to a unique output value. If the function f(x) = x^2 is defined on the set of real numbers, then it is not injective because multiple input values (e.g. 2 and -2) map to the same output value (4). Therefore, the function f(x) = x^2 is not injective. **

One example of a function that is injective but not surjective is f(n) = n + 1. This function takes a natural number n and maps it to n + 1. It is injective because different natural numbers will always map to different results, but it is not surjective because there is no natural number that maps to 1. **

One way to prove that function f is injective if function g is injective is to show that for any two distinct inputs x1 and x2, the outputs f(x1) and f(x2) are also distinct. Since g is injective, we know that g(x1) and g(x2) are distinct, and we can use this property to show that f is injective as well. Specifically, we can use the fact that g(f(x1)) = g(f(x2)) implies f(x1) = f(x2), and since g is injective, this implies x1 = x2. Therefore, f is injective. **

One example of a function that is injective but not surjective is the function f(n) = n + 1. This function takes a natural number n and maps it to the natural number n + 1. It is injective because if f(a) = f(b), then a + 1 = b + 1, so a = b. However, this function is not surjective because there is no natural number n such that f(n) = 1, since the function only outputs numbers greater than or equal to 2. Therefore, f: N -> N is injective but not surjective. **

To show that if f and g are injective, then gf is also injective, we can use the definition of injective functions. An injective function is one where distinct inputs map to distinct outputs. So, if f and g are injective, then for any distinct inputs x1 and x2, f(x1) ≠ f(x2) and g(y1) ≠ g(y2) for any distinct outputs y1 and y2. Now, consider the composition gf. If gf(x1) = gf(x2), then f(x1) = f(x2), which implies x1 = x2 by the injectivity of f. Therefore, gf is also injective. **

A matrix is injective if its columns are linearly independent, meaning that the only solution to the equation Ax=0 is x=0. A matrix is surjective if its columns span the entire codomain, meaning that for every element in the codomain, there exists at least one vector x such that Ax=b. A matrix is bijective if it is both injective and surjective, meaning that it has a unique solution for every element in the codomain. **

To determine if a mapping is surjective or injective, we need to look at the properties of the mapping. Please provide the specific mapping you would like me to analyze. **

The first mapping is injective because each element in the domain is mapped to a unique element in the codomain. The second mapping is surjective because every element in the codomain is mapped to by at least one element in the domain. **