I need to prove a proposition that states that a product $\sigma$-algebra can be generated by a certain set, only I can't figure out why a projection's inverse is well defined. I will give the definition so the notation is clear.
Let $(\Bbb{X}_n)$ be a sequence of non-empty sets. The product set $$\Bbb{X} = \prod_{n=1}^\infty \Bbb{X}_n$$ consists of sequences in which the $n$-th element belongs to $\Bbb{X}_n$. For every $x = (x_1, x_2, ...) \in \Bbb{X}$ we define the projections $$\pi_n: \Bbb{X} \rightarrow \Bbb{X}_n \quad , \quad \pi_n: x \mapsto x_n$$ For every $n \in \mathbb{N},$ let $\Sigma_n$ be a $\sigma$-algebra on $\Bbb{X}_n$. The $\textbf{product}$ $\textbf{$\sigma$-algebra}$ on $\Bbb{X}$ is defined by $$\otimes_{n=0}^\infty\Sigma_n = \sigma{(\{\pi_{n}^{-1}(A_n) : A_n \in \Sigma_n, \quad n \in \mathbb{N}}\})$$
So I'm asking $\textbf{why does the projection $\pi_n$ have a well defined inverse?}$ It looks to me as though the inverse can produce more than one output for a given input.
Here $\pi_n^{-1}$ means the inverse image, not the inverse function. If $f:X\to Y$ is a function and $A\subseteq Y$, then the inverse image $f^{-1}(A)$ is defined as $\{x\in X:f(x)\in A\}$.