Function preserving exponentiation [duplicate]

Solution 1:

First off, the constant functions $-1$, $0$ (where $0^0 = 0$), and $1$ satisfy the functional equation. Let us now find the remaining continuous functions that satisfy the equation.

According to Martin Nicholson's comment, for all $a$, $x$, $y$: $$f(a)^{f(xy)} = f(a^{xy}) = f(a^x)^{f(y)} = f(a)^{f(x)f(y)}.$$ Therefore if there is some $a$ for which $1 \ne f(a) > 0$ then $f(xy) = f(x)f(y)$. By this answer, $f(x)$ must then be of the form $x^c$ for some constant $c$. So: $$(a^b)^c = f(a^b) = f(a)^{f(b)} = (a^c)^{b^c}$$ which implies that $c \in \{0, 1\}$. If $c$ is $0$ then $f = 1$; otherwise $f$ is the identity function.

Now we show that such an $a$ mentioned above exists. Note that $f$ preserves power towers: $$f\left(x^{x^\ldots}\right) = f(x)^{f\left(x^{x^\ldots}\right)}$$ so $$f\left(x^{x^\ldots}\right) = f(x)^{f(x)^\ldots}.$$ As power towers are defined for exactly those $x \in I := \left[e^{-e}, e^\frac{1}{e}\right]$ and as $f\left(x^{x^\ldots}\right)$ is defined for any $x \in I$, $f$ maps this interval to itself. If $f$ were $1$ on this interval then the functional equation would require $f$ to be $1$ everywhere. Thus if $f \ne 1$ then there must be some $a \in I$ for which $f(a) \in I$, $f(a) \ne 1$, which gives the required $a$.

In summary, the continuous functions that satisfy the functional equation are $-1$, $0$ (if $0^0 = 0$), $1$, and the identity.