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Math Help - Projections

  1. #1
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    Projections

    Hi guys,

    The question I'm doing states,

    "Let i_1, i_2, \ldots, i_n be indices in I, and assume that for each i \in I, we are given U_i \subseteq X_i. Show that \bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r}) can be described as a product of subsets of X_i's"

    I just want to see if my thinking here is correct.

    We have that \pi^{-1}_{i} is given by the projection onto the ith factor. So, \pi^{-1}_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}, for instance. We have the following fact for some U_1\subseteq X_1, U_2 \subseteq X_2,

    \pi^{-1}_1(U_1) = U_1 \times X_2

    where \pi_1 : U_1 \times U_2 \rightarrow U_1

    Now, I know I have to use this to prove the above question, I'm just stuck on how I would represent \pi^{-1}_{i_r}(U_{i_r}) using the above, and how to then manipulate the intersections. There is a further property that

    \pi^{-1}_1(U_1) \cap \pi^{-1}_1(U_2) = U_1 \times U_2,

    which perhaps may come in useful, although I haven't seen any use for it.

    Thanks in advance,

    HTale.
    Last edited by HTale; February 8th 2009 at 05:07 PM.
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  2. #2
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    Quote Originally Posted by HTale View Post
    \pi^{-1}_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}, for instance. We have the following fact for some U_1\subseteq X_1, U_2 \subseteq X_2

    It should be
    \pi_{i_1} : U_{i_1} \times U_{i_2} \times \cdots \times U_{i_n} \rightarrow U_{i_1}.

    Let S be a subbasis for the product topology you mentioned. Then,

    S = \{\pi^{-1}_{i}(U_{i}) : U_{i} is open in X_{i}, i \in I \}.

    Since the all finite intersection of elements of S forms a basis for (X, T),
    \bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r}) is a basic open set for a topology T. If we denote a basic open set for a product topology T using a product form,

    \bigcap_{1\leq r \leq n} \pi^{-1}_{i_r}(U_{i_r}) = \prod _{i \in I} A_{i}, where A_{i} = U_{i} for i = i_1, i_2, .., i_n in  I and A_{i} = X_{i} otherwise.
    Last edited by aliceinwonderland; February 9th 2009 at 03:32 AM.
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