Final answer:
To prove that f is bijective, we need to show that it is both injective (one-to-one) and surjective (onto).
Step-by-step explanation:
To prove that f is bijective, we need to show that it is both injective (one-to-one) and surjective (onto).
Let's start with the forward direction - assume f is bijective. To show that f(A/C) = B/f(C), we need to prove that for every element y in B/f(C), there exists a corresponding element x in A/C such that f(x) = y.
Now, let's move to the reverse direction - assume f(A/C) = B/f(C). In this case, we need to prove that f is one-to-one and onto.
To prove injectivity, we assume that f(x1) = f(x2) for some x1 and x2 in A/C. This implies that x1 and x2 must be equal, as otherwise, they would have different images in B/f(C).
To prove surjectivity, we need to show that for every element y in B, there exists an element x in A such that f(x) = y. By using the fact that f(A/C) = B/f(C), we can find x in A/C that maps to y in B. Since A/C is a subset of A, we can conclude that f is onto.