Matthew, but what is \\(g\\) and \\(\lambda\\) in (2)?

I tried the following, where \\(F\\) stands for \\((-)^B\\):

- let B = 1, then we get \\(X^1 \cong X\\) and \\(F(f): x \mapsto (f(x)) \\)

- let B = 2, then we get \\(X^2 \cong X \times X\\) and \\(F(f): (x, x') \mapsto (f(x), f(x'))\\), or in other words \\(F(f): X \times X \to Y\times Y\\), which in turn \\(Y \times Y = Y^2\\)

- for arbitrary B (even not finite), we get \\(X^B \cong (X \times X \times ... \times X)_B \\), and therefore \\(F(f): (x_1, x_2, ...) \mapsto (f(x_1), f(x_2), ...)) \\)

However, at the moment I don't know whether this makes sense and how to make it look better...

I tried the following, where \\(F\\) stands for \\((-)^B\\):

- let B = 1, then we get \\(X^1 \cong X\\) and \\(F(f): x \mapsto (f(x)) \\)

- let B = 2, then we get \\(X^2 \cong X \times X\\) and \\(F(f): (x, x') \mapsto (f(x), f(x'))\\), or in other words \\(F(f): X \times X \to Y\times Y\\), which in turn \\(Y \times Y = Y^2\\)

- for arbitrary B (even not finite), we get \\(X^B \cong (X \times X \times ... \times X)_B \\), and therefore \\(F(f): (x_1, x_2, ...) \mapsto (f(x_1), f(x_2), ...)) \\)

However, at the moment I don't know whether this makes sense and how to make it look better...