Show that: $[(X_1,T_1) \times (X_2, T_2)] \times (X_3,T_3) \cong (X_1,T_1) \times (X_2, T_2) \times (X_3,T_3) $.

I see that the first space is of the form $$\{((a,b),c): a,b \in X_1 \times X_2 \land c \in X_3\}$$ while the second is of the form $$\{(a,b,c): a \in X_1 \land b \in X_2 \land c \in X_3\}$$ But I'm not sure how I would create a bijective function between the two spaces besides just splitting up the first coordinate in the first set to individual coordinates in the second set.

  • $\begingroup$ So what's wrong with just splitting up the first coordinate? $\endgroup$ – Anon Jun 11 '16 at 12:28
  • $\begingroup$ Perhaps you mean to show $(X_1\times X_2)\times X_3$ and $X_1\times (X_2\times X_3)$ are homeomorphic? If so, that would be an associative property for the product space topology. $\endgroup$ – hardmath Jun 11 '16 at 13:29

The map you describe is the required bijection. The inverse is also clear.

Now use that a map into a product is continuous iff the composition of that map with all the projections of that product is continuous, for both of these bijections, and show they're both continuous that way.


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