# Difference between $T(A) = A$ and $T^{-1}(A) = A$

I am a bit confused about the title.

Let $$T:X\rightarrow X$$ be a map where $$X \subset \mathbb{R}^n$$. Let $$A\subset X$$.

I know that $$T^{-1}(A) = \{x\in X: T(x)\in A\}.$$

and if $$A$$ is $$T$$-invariant, we have $$T(A) = A.$$ In this case, $$A$$ is a bit like an attractor.

My questions are:

1. Suppose $$T^{-1}(A)=A$$, can we say $$T(A) = A$$? or we can just say $$T(A)\subset A$$?
2. Suppose $$T(A) = A$$, can we say $$T^{-1}(A) = A$$?

I think 2. is not correct; because if $$A$$ admits a basin of attraction of $$A$$, say $$V$$, such that $$A\subset V\subset X$$, then it could be possible that $$A\subset T^{-1}(A)$$.

Let $$X=\mathbb{R}$$, and let $$T:X\to X$$ be given by $$T(x)=x^2$$.
For the first one, if $$A=X$$, then $$T^{-1}(A)=A$$, but $$T(A)=[0,\infty)$$ which is a proper subset of $$A$$.
For the second one, if $$A=[0,\infty)$$, then $$T(A)=A$$, but $$T^{-1}(A)=X$$ which properly contains $$A$$.