Composition of weak one way function is not a strong one way function Given $f(x)$, a weak one-way permutation, how to prove that $f^T(x)$ is not a strong one-way function? Here $f^T$ denotes 
$T$ times self composition of $f$ and $T$ is a polynomial in input length.
Obviously, composing $T$-times a weak OWP can give a strong OWF: simply consider the case where $f$ is already a strong OWP (hence in particular a weak OWP). Now, what you want is to show that it is not necessarily the case; i.e., you want a function $f$ which is a weak OWP, but such that $f^T$ is not a strong OWF (i.e., an example where security does not "amplify through composition").

I'll give you such a candidate; I let you prove that it satisfies what you want, that is a quite simple exercise. Let $g: I \mapsto I$ be an arbitrary (strong or weak) OWP, and let $J$ be some arbitrary set such that $|J| = |I|$, and $I$ is disjoint from $J$. Consider the following function $f$:

$f:I \cup J \mapsto I \cup J$ is defined as follows: $f(x) = g(x)$ if $x \in I$, and $f(x) = x$ if $x\in J$.

Claim: $f$ is a permutation (that's quite obvious), and also a weak-OWP (you should be able to check that easily). Yet, $f^T$ is still a weak-OWP, but not a strong OWF. Can you see why?

(if you're stuck in the analysis, just ask for more hints in the comment section)