Showing a CBC-MAC is not preimage resistant with a proof So hypothetically I have a an arbitrary block cipher operating in CBC-MAC mode that makes use of a public and static $IV$ as well as a static key $K$.
I want to be able to that this won't be preimage resistant using a proof for an arbitrary single-block message $P$ that hashes to a given digest $T$. So far I've gotten as far as $$T = E(P \oplus IV, K)$$ $$D(T, K) = P \oplus IV$$ $$P = D(T, K) \oplus IV$$
I can't for the life of me figure out how to move from this to something that will show that the operation is not preimage resistant. As far as I understand if $K$ and $IV$ are fixed it should be computationally infeasible to find another $T_x$ such that $$D(T,K) \oplus IV = D(T_x, K) \oplus IV$$
Ideally I'd want to be able to extend this to show that for $P$ of any length I could find a $T_x$.
According to wikipedia's page on CBC-MAC:

If the block cipher used is secure (meaning that it is a pseudorandom permutation), then CBC-MAC is secure for fixed-length messages.[1] However, by itself, it is not secure for variable-length messages. Thus, any single key must only be used for messages of a fixed and known length.

Your case presents no such vulnerability, since you're hashing a fixed number of blocks ($1$, to be exact). If you were to compute the CBC-MAC of a variable-length message you could perform such a preimage attack.