The equality of a finite intersection with a cartesian product. So, I'm currently going through Munkres' Topology, and I have arrived at his introduction to the product topology on arbitrary cartesian products. 
In his introduction, he defines an arbitrary cartesian product as follows:

Let $\{A_\alpha\}_{\alpha \in J}$ be an indexed family of sets. The Cartesian Product of this family is the set of all functions $x: J \to \bigcup_{\alpha \in J} A_\alpha$ such that $x(\alpha) \in A_\alpha$ for each $\alpha \in J$.

Then, if $M :=\Pi_{\alpha \in J} X_\alpha$ is a cartesian product of topological spaces $X_\alpha$, he defines the product topology on $M$ as the topology generated by the subbasis $$S = \bigcup_{\beta \in J} S_\beta$$
where $S_\beta = \{\pi_{\beta}^{-1}(U_\beta): U_\beta$ is open in $X_\beta\}$. Here, $\pi_\beta$ is the projection $M \to X_\beta$ mapping each element of $M$ to its $\beta$th coordinate (i.e. $\pi_\beta(x) = x(\beta)$). 
He goes on to show that if $\beta_1, \ldots, \beta_n$ is a finite collection of elements of $J$ and $U_{\beta_i}$ is open in $X_{\beta_i}$, then \begin{equation} \pi_{\beta_1}^{-1}(U_{\beta_1}) \cap \ldots \cap \pi_{\beta_n}^{-1}(U_{\beta_n}) = \Pi_{\alpha \in J} U_\alpha \ \ \ \ (1)\end{equation} where $U_\alpha = X_\alpha$ if $\alpha \neq \beta_1, \ldots, \beta_n$ (I can provide more detail on how he comes to claim this if need be).
Now, on the left hand side, we notice that $\pi_{\beta_i}^{-1}(U_{\beta_i}) = \{ x \in M: \pi_{\beta_i}(x) \in U_{\beta_i}\}$. In particular, an element of the LHS of $(1)$ is a function $J \to M$. However, by the given definition of the cartesian product, the elements of $\Pi_{\alpha \in J} U_\alpha$ are functions $J \to \bigcup_{\alpha \in J} U_\alpha$. I've always understood that one of the criteria that functions must satisfy in order to be equal is that they must have the same codomain. So, if $M$ is not necessarily equal to $\bigcup_{\alpha \in J} U_\alpha$, how can a function be in both in the LHS and the RHS of $(1)$? From what I see, the functions in the LHS and RHS have different codomains. I know in the big scheme of things, this is probably a really minor issue, but the question has been itching at me for a little while, and I would love if someone could shed some light on something that perhaps I'm failing to see. Thank you. 
Edit: In my last paragraph, I accidentally write $J \to M$ when I mean $J \to \bigcup_{\alpha \in J} X_\alpha$, and it's this codomain that is not necessarily equal to $\bigcup_{\alpha \in J} U_\alpha$
 A: You are certainly aware that $A'\subset A$, $B'\subset B$, $C'\subset C$ implies $A'\times B'\times C'\subset A\times B\times C$, because every ordered triple $(x,y,z)$ with $x\in A'$, $y\in B'$, $z\in C'$ is also a triple with $x\in A$, $y\in B$, $z\in C$. But when we view $A\times B\times C$ (and similarly for $A'\times B'\times C'$) as the set of maps $f\colon \{1,2,3\}\to A\cup B\cup C$ with the property $f(1)\in A$ and $f(2)\in B$ and $f(3)\in C$ instead of a set of triples, then your objections kick in already in the case of finite products.
So what we do (and what perhaps should have been specified in your text) is that in set theory we care less about domain and co-domain of a function, i.e., in set theory function usually does not mean a triple $(A,B,G_f)$ of a set $A$, a set $B$, and a set $G_f\subseteq A\times B$ with the property that for each $a\in A$ there exists exactly one $b\in B$ such that $(a,b)\in G_f$. Instead, we tend to remove all ballast and identify the function with $G_f$ alone. Clearly, $A$ can be recovered as the set of all first components of elements of $G_f$, whereas $B$ cannot be completely recovered. An arbitrary set $G_f$ is then considered a function if 


*

*If $z\in G_f$ then $z$ is a (Kuratowski) pair $z=(x,y)$ of some sets $x$ and $y$

*if $(x,y)\in G_f$ and $(x,y')\in G_f$ then y=y'$.


Thus you may view $G_f$ as a function with co-domain the all-class and domain some set - or even as a partial function from the all-class to the all-class and perhaps even allow $G_f$ to be a proper class.
In fact, the very axioms of ZFC make use of this in the Axiom Schema of Replacement: If $\Phi$ is a class function, i.e., a two place predicate $\Phi(a,b)$ with the property that $\forall a,b,c\colon \Phi(a,b)\land \Phi(a,c)\to b=c$, then for any set $A$, there exists a set $\Phi[A]$ with $x\in\Phi[A]\iff \exists a\in A\colon \Phi(a,x)$.
Without specifying the co-domain, we lose the ability to speak of surjectivity, and when using composing partial functions we no longer automatically have that $g\circ f$ has the same domain as $f$. Therefore, this simplified(?) set-theoretic view of the notion of function is often much less useful than taking domain and co-domain explicitly into the definition.
To cut a long story short, you could forget about this and instead write the problematic statement as

$$\pi_{\beta_1}^{-1}(U_{\beta_1}) \cap \ldots \cap \pi_{\beta_n}^{-1}(U_{\beta_n}) = \operatorname{im}\iota$$
  where $\tilde\iota\colon \Pi_{\alpha \in J} U_\alpha\to M$ is the canonical embedding $f\mapsto \iota\circ f$ induced by the inclusion map $\iota\colon\bigcup_\alpha U_\alpha\to\bigcup_\alpha X_\alpha$.

This is on the other hand perhaps much less appealing.
A: I haven't read Munkres' book, but from what you have written, it seems to me that he just does a slight abuse of notation. Any function $x : J \to \bigcup_{\alpha \in J}U_\alpha$ can be viewed as a function with codomain $\bigcup_{\alpha \in J}X_\alpha$, simply define $\tilde{x} : J \to \bigcup_{\alpha \in J}X_\alpha$ via $\tilde{x}(\alpha) := x(\alpha) \in U_\alpha \subset X_\alpha$. 
If you want to be precise, the construction you mention at first is how we define a product for a given family of spaces $(X_\alpha)_{\alpha \in J}$. Now, if we have subsets $A_\alpha \subset X_\alpha$ for each $\alpha$, we can define
$$
\prod_{\alpha \in J} A_\alpha := \left\{x \in \prod_{\alpha \in J}X_\alpha : x(\alpha) \in A_\alpha\right\}.
$$ 
You can check that this subset provided with the subspace topology is homeomorphic to the original definition of $\prod_{\alpha \in J} A_\alpha$ as the product of each space $A_\alpha$, hence the abuse of notation.
A: Well the co-domain of the elements of the left side of (1) is a subset of the co-domain of the right side so there is and obvious imbedding of these left side elements into elements with the codomain of the right side so that your equality holds .
    Another way of looking at it is to have what you call function to be called mapping and have function mean the graph of a mapping . The graph still has a domain (and a range ) but no codomain . Then define the product as a collection of functions instead of mappings . But technically with your notation you are right
