$f^*(U_1 \times ... \times U_k) = \bigcap_{i=1}^k f^*_i(U_i)$

Question

I’m trying to prove this result and I would really appreciate if you could give some feedback in my proof.

Result: Let $A,A_1,...,A_k$ be sets, for some positive integer $k$, let $f: A \rightarrow A_1 \times ... \times A_k$ be a function and let $U_i \subseteq A_i$ for each $i \in \{1,...,k\}$. Then $$f^*(U_1 \times ... \times U_k) = \bigcap_{i=1}^k f^*_i(U_i).$$

I will be using the map $\pi_i:A_1 \times ... \times A_k \rightarrow A_i$ defined by $\pi_i((x_1,...,x_k))=x_i$ for all $(x_1,...,x_k) \in A_1 \times ... \times A_k$ (the projection map). In addition, I will make use of the functions $f_i: A \rightarrow A_i$ defined by $f_i = \pi_i \circ f$ for all $i \in \{1,...,k\}$ (the component functions of $f$).

Proof: In order to show that $f^*(U_1 \times ... \times U_k) = \bigcap_{i=1}^k f^*_i(U_i)$, we have to show that each of the sets is a subset of the other.

Let $a \in f^*(U_1 \times ... \times U_k)$. We have that $f(a) \in U_1 \times ... \times U_k$. Then $f(a)=(u_1,...u_k)$ with $u_i \in U_i$ for each $i \in \{1,...,k\}$. By the definition of component functions, we see that $u_i = (\pi_i \circ f)(a)=f_i(a)$ for all $i \in \{1,...,k\}$. Hence $f_i(a) \in U_i$ for all $i \in \{1,...,k\}$. By definition, we have that $a \in f^*_i(U_i)$ for all $i \in \{1,...,k\}$, so $a \in \bigcap_{i=1}^k f^*_i(U_i)$. Therefore $f^*(U_1 \times ... \times U_k) \subseteq \bigcap_{i=1}^k f^*_i(U_i)$.

Now, let $b \in \bigcap_{i=1}^k f^*_i(U_i)$. We have that $b \in f^*_i(U_i)$ for all $i \in \{1,...,k\}$. By definition, $f_i(b) \in U_i$ for all $i \in \{1,...,k\}$. This implies that $(f_1(b),...,f_k(b)) \in U_1 \times ... \times U_k$, so $f(b) \in U_1 \times ... \times U_k$. By definition, we conclude that $b \in f^*(U_1 \times ... \times U_k)$. Therefore $\bigcap_{i=1}^k f^*_i(U_i) \subseteq f^*(U_1 \times ... \times U_k)$.

With this, we conclude that $f^*(U_1 \times ... \times U_k) = \bigcap_{i=1}^k f^*_i(U_i)$.

$\square$

Thank you very much!

Answer 1

A more simplified proof: \begin{align} f^*(U_1 \times \cdots \times U_k) &= \{a \in A : f(a) \in U_1 \times \cdots \times U_k\} \\ &= \{a \in A : (f_1(a),\dots,f_k(a)) \in U_1 \times \cdots \times U_k\} \\ &= \{a \in A : f_i(a) \in U_i \textrm{ for all } i \in \{1,\dots,k\}\} \\ &= \{a \in A : a \in f^*(U_i) \textrm{ for all } i \in \{1,\dots,k\}\} \\ &= \bigcap_{i=1}^k f^*(U_i). \end{align}

Answer 2

The asterisk notation is highly uncanonical and extremely strange to be used in a setting -- such as that of set theory -- where there is such a natural, common-sensical and well-established tradition of denoting inverse images with $\bullet^{-1}$, tradition which I dare recommend to you as well.

Given a family $B$ of sets indexed by a nonempty index set $I \neq \varnothing$ (the assumption of nonemptiness serves to prevent the case of empty intersections, which the particular axiomatic system I am an adept of does not handle), a set $A$ and a family of maps $f \in \displaystyle\prod_{i \in I}\mathrm{Hom}_{\mathbf{Ens}}(A, B_i)$ (which simply means that at each index $i \in I$ the component $f_i$ of the family $f$ is a map from $A$ to $B_i$; $\mathbf{Ens}$ is the name I give to the category of sets), let us denote by $g \colon A \to \displaystyle\prod_{i \in I}B_i$ the direct product of family $f$ in restricted sense (also called by some diagonal product), in other words the unique map $h \colon A \to \displaystyle\prod_{i \in I}B_i$ such that $p_i \circ h=f_i$ for each index $i \in I$, where $p_i$ are the canonical projections of the cartesian product of family $B$. Your objective is to prove that for any family $X \in \displaystyle\prod_{i \in I}\mathscr{P}(B_i)$ of subsets (again, this simply means that at each index $i \in I$ the component $X_i$ of family $X$ is a subset of $B_i$) the relation: $$g^{-1}\left[\prod_{i \in I}X_i\right]=\bigcap_{i \in I}f_i^{-1}[X_i]$$ is valid.

A perhaps even more elegant and immediate proof is available once you observe the fact that: $$\prod_{i \in I}X_i=\bigcap_{i \in I}p_i^{-1}[X_i],$$ since by taking inverse images via $g$ in the above relation you easily obtain: $$\begin{align*} g^{-1}\left[\prod_{i \in I}X_i\right]&=g^{-1}\left[\bigcap_{i \in I}p_i^{-1}[X_i]\right]\\ &=\bigcap_{i \in I}g^{-1}\left[p_i^{-1}[X_i]\right]\\ &=\bigcap_{i \in I}\left(p_i \circ g\right)^{-1}[X_i]\\ &=\bigcap_{i \in I}f_i^{-1}[X_i]. \end{align*}$$