I have a question w.r.t. notation that appears dealing with product sets. Let $T$ be any index set and $X$ and $Y$ be any two sets. Then $X^T = \{f:T\to X\} $ and $Y^T = \{g:T\to Y\}$. Let $A\subset Y^T$, I am interested how shall I denote $$ Z:=\{(f,g)\in (X\times Y)^T|g\in A\}. $$ Clearly, $X^T\times A$ is isomorphic to $Z$ but $Z\neq X^T\times A$. I am looking for the shorthand notation of $Z$ since I am dealing with many $A$'s and thus defining it each time is not quite a solution.

I thought of $X\parallel A$ since it's a kinda parallel product, but I'm not sure that's the best choice.

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    $\begingroup$ I don't think your definition of $Z$ makes sense, because an element of $(X \times Y)^T$ is not an ordered pair. Perhaps you meant $Z$ to be the set of functions $h \in (X \times Y)^T$ such that composition with the coordinate projection onto $X$ yields a function in $A$? $\endgroup$ – Trevor Wilson Mar 4 '13 at 17:11
  • $\begingroup$ @TrevorWilson: thanks, that's what I meant - rather the projection onto $Y$ shall yeild a function in $A$, though $\endgroup$ – Ilya Mar 5 '13 at 8:12

As pointed out by Trevor, you probably meant

$$Z = \Big\{ f \in (X \times Y)^T\ \Big|\ \pi_Y \circ f \in A \Big\}.$$

As you observed, $Z$ is isomorphic to $X^T\times A$. Why not just use that?

Let $\xi : X^T \times Y^T \to (X\times Y)^T$ be defined as $$\xi(f,g)(t) = \big(f(t), g(t)\big)$$ or in different notation $$\xi(f,g) = t \mapsto \big(f(t), g(t)\big).$$

Then, you can set $$Z = \vec\xi(X^T \times A).$$

I hope this helps ;-)

  • $\begingroup$ This certainly helps, thanks :) I'll wait if more answers come, and if not I'll accept yours $\endgroup$ – Ilya Mar 5 '13 at 8:15

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