Let $X$ be compact and $f:X \to X$ an expanding map. It is well known that $f$ admits a unique fixed point since $f^{-1}$ has to be a contraction. Ok.
Now, consider this: take $U \subset X$ open arbitrary, and $n \in \mathbb{N}$ such that $f^n(U)=X$. In this fashion there is a copy of $f^{-n}(U)$ inside $U$, a copy of $f^{-2n}(U)$ inside $f^{-n}(U)$, and so on.
Take $\bigcap_{j \in \mathbb{N}} \left(U \cap f^{-nj}(U)\right)$ so why is not in there a fixed point for $f$?
Edit: It is straightforward that $f$ can not admit two fixed points since it would contradict the definition of expanding, my question is why is that reasoning misleading? Thanks in advance for taking your time to think along. Cheers!
You are using the inverse, but a expanding map on a compact space is not invertible. If it is surjective, like $2 \times Id$ on $\mathbb{T}^2$, every point has more than one preimage (4, in this case). You can, however, work with pre images. If $U$ is connected and open, $f^{-n}(U)$ has several connected components $V_i$, each one of them satisfying $f(V_i) = U$. Regardless of this, it is not true that there is a connected component of $f^{-n}(U)$ inside $U$ for some $n \in \mathbb{N}$.
Example: take $f = 2 \times Id: \mathbb{T}^2 \to \mathbb{T}^2$ and let $U = (0.4, 0.6) \times (0.4, 0.6)$. Then $$f^{-1}(U) = (0.2, 0.3) \times (0.2, 0.3) \cup (0.2, 0.3) \times (0.7, 0.8) \cup (0.7, 0.8) \times (0.2, 0.3) \cup (0.7, 0.8) \times (0.7, 0.8),$$
and $f^{-1}(U)$ and $U$ are disjoint.