Topology: An Inquiry-Based Approach

Given topological spaces \((X_1, \tau_1)\) and \((X_2, \tau_2)\text{,}\) we define \(\pi_1: X_1 \times X_2 \to X_1\) and \(\pi_2: X_1 \times X_2 \to X_2\) by \(\pi_1((x,y)) = x\) and \(\pi_2((x,y)) = y\text{.}\) These functions \(\pi_1\) and \(\pi_2\) are the projections of \(X_1 \times X_2\) onto \(X_1\) and \(X_2\text{,}\) respectively. These projection functions can help us determine when a function \(f\) from a topological space \(Y\) to \(X_1 \times X_2\) is continuous.

The same argument as in Activity 20.6 shows that \(\pi_2\) is also a continuous function. In general, if \(X = \Pi_{i=1}^n X_{i}\) is a finite product of topological spaces, then the projection \(\pi_{k}: X \to X_{k}\) is a continuous function for each \(k\text{,}\) where \(\pi_k((x_1,x_2, \ldots, x_n)) = x_k\text{.}\)

So each basic open set is a finite intersection of sets of the form \(\pi_i^{-1}(O_i)\) where \(O_i\) is open in \(X_i\text{.}\) When this happens, we call the collection of sets of the form \(\pi_i^{-1}(O_i)\) a subbasis of the topology.

This set is a subbasis for the product topology on \(X\) (the verification of this is left to Exercise 1).

For reasons we won’t go into, the product topology is preferred to the box topology for infinite products (many important theorems that hold for finite products will not hold for infinite products using the box topology, but will hold using the product topology). However, the product topology and the box topology are the same for finite products, and since we won’t consider infinite products here we will not worry about the distinction. For our purposes we will use the terms “box topology” and “product topology” interchangeably.

As we have discussed before, it can often be easier to define a topology using a basis or subbasis than it is to describe all of the sets in the topology. As we might expect, since the continuity of a function can be determined by the inverse image of basis elements, the continuity of a function can also be determined by the inverse image of subbasis elements.

Now suppose that \(X_1\text{,}\) \(X_2\text{,}\) and \(Y\) are topological spaces, and that \(f: Y \to X_1 \times X_2\) is a function. Then \(\pi_1 \circ f\) maps \(Y\) to \(X_1\) and \(\pi_2 \circ f\) maps \(Y\) to \(X_2\text{.}\) Since the composition of continuous functions is continuous, we can see that if \(f\) is continuous so are \(\pi_1\circ f\) and \(\pi_2 \circ f\text{.}\) To determine if \(f\) is a continuous function, it would be useful to know if the converse is true. A key idea in the proof is the result of Exercise 9 that if \(R\text{,}\) \(S\text{,}\) and \(T\) are sets, and \(g: R \to S\) and \(h : S \to T\) are functions, then \((h \circ g)^{-1}(O) = g^{-1}(h^{-1}(O)\) for any subset \(O\) of \(T\text{.}\)

Now we can use projections to determine when functions to product spaces are continuous.