When is a $(\prod_{\alpha \in J} X_\alpha \to X)$-function continuous?

Question

I know that a function $X \to \prod_{\alpha \in J} X_\alpha$ from a topological space into a space with the product topology is continuous if and only if each component function $f_\alpha : X \to X_\alpha$ is continuous. My question, is there a main theorem for functions of the type $\prod_{\alpha \in J} X_\alpha \to X$?

Answer 1

There is no main theorem for functions of that type (although you might plausibly see some theorems for specific cases -- of course, that makes this question hard to answer with a strong "no").

In particular, the theorem that you might expect to exist would be something like: if for each $a_0 \in J$ and each sequence of points $(x_a)_{a \in J \setminus \{a_0\}}$, the composition $$ X_{a_0} \times \prod_{a \in J\setminus \{a_0\}} \{x_a\} \to \prod_{a \in J} X_a \to X $$ is continuous, then $\prod_{a \in J} X_a \to X$ is also continuous.

In fact, this statement is not true. To see this consider the following subset of $[0, 1]^3$: it includes the lines $\{(0, y, 1) \mid y \in [0,1]\}$ and $\{(x, 0, 1) \mid x \in [0,1]\}$, and it also includes the line $\{(x, x, 0) \mid x\in (0, 1]\}$. Now, for each $x \in (0, 1]$, the straight line from $(x, 0, 1)$ to $(x, x, 0)$ is in the set, as is the straight line from $(0, x, 1)$ to $(x, x, 0)$.

With this description you can check that for each $(x, y) \in [0,1]^2$ there is a unique $z \in [0,1]$ so that $(x, y, z)$ is in the set; hence, the set is the graph of some function $f: [0,1]^2 \to [0,1]$. However, you can check that for each $x \in [0,1]$, the functions $y \mapsto f(x, y): [0, 1] \to [0,1]$ and $y \mapsto f(y, x): [0,1] \to [0,1]$ are continuous.

On the other hand, there is an object for which you get a nice theorem classifying outgoing maps in terms of components. It is not the product -- it is the disjoint union.