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Are Projection Mappings considered Quotient Maps?

  1. Apr 27, 2015 #1
    The book I am using for my Introduction to Topology course is Principles of Topology by Fred H. Croom.

    Problem:
    Prove that if ##X=X_1\times X_2## is a product space, then the first coordinate projection is a quotient map.

    What I understand:
    • Let ##X## be a finite product space and ## x=(x_1,x_2,...,x_n)## a point in ##X##. The function ##\pi_i:X\rightarrow X_i## defined by ##\pi_i(x)=x_i## is called the projection map on the ##i##th coordinate.
    • Let ##X## and ##Y## be two topological spaces and ##f:X→Y## a surjective map. The map ##f## is said to be a quotient map provided a subset ##U## of ##Y## is open in ##Y## if and only if ##f^{−1}(U)## is open in ##X##. Equivalently, subset ##U## of ##Y## closed in ##Y## if and only if ##f^{−1}(U)## closed in ##X##.
    Rough attempt:

    To show ##\pi_1:X\rightarrow X_1## is a quotient map, I need to show that (1) ##\pi_1## is surjective, (2) ##\pi_1## is continuous, and (3) for some ##U_1\subset X_1, \pi_1^{-1}(U_1)## open in ##X \Rightarrow U_1## is open in ##X_1##.
    ##\pi_1## is clearly onto since ##\forall x\in X, \exists x_1\in X_1: \pi_1(x)=x_1.## Let ##O_1## be open subset of ##X_1##. Then ##\pi_1^{-1}(O_1)=\{x=(x_1,x_2)\in X:x_1\in O_1\}=O_1\times X_2##. Thus ##\pi_1^{-1}(O_1)## is open in ##X##. Thus ##\pi_1## is continuous. The third part is what I am not sure of. Since ##\pi_1## is surjective, we have ##\pi_1(\pi_1^{-1}(V_1))=V_1, \forall V_1\subset X_1##. Assume now that ##W=\pi_1^{-1}(O_1)## is an open subset of ##X##. My question is, how would I conclude that ##\pi_1(W)=O_1## is indeed open. Can I just assume or is there something I am missing? I'm sure I am missing something.


    Out of curiosity, could I generalize this and say if ##X=\prod_1^nX_i## if a finite product space, then the first coordinate projection is a quotient map? How about the ##i##th coordinate projection? I want so say yes. I can easily prove that ##\pi_i:X\rightarrow X_i## is continuous using the same method I did for continuity above. Anywho, that's probably a different story for a different time.


    Sorry for the long read. If I need to clarify on anything I presented, let me know. I sincerely thank you for taking the time to read this post. I greatly appreciate any assistance you may provide.
     
  2. jcsd
  3. Apr 27, 2015 #2

    Dick

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    No, you can't just assume it. You want to think about the basis for the product topology. It's all sets of the form ##O_1 \times O_2## where ##O_1## and ##O_2## are open in ##X_1## and ##X_2## respectively. Can you use that to show that if ##W## is an open set in ##X_1 \times X_2## then if ##x \in W## then there is a basis set ##O_1 \times O_2## that contains ##x## and is also a subset of ##W##? That would do it, yes?

    And sure, there is nothing special about the first coordinate.
     
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