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[topology] compact, locally connected, quotient topology

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Let X be a compact and locally connected topological space. Prove that by identifying a finite number of points of X, one gets a topological space Y that is connected for the quotient topology.

    2. Relevant equations
    The components of a locally connected space are open.

    3. The attempt at a solution
    And thus the set of components [itex]\mathcal C[/itex] is an open covering of X and since the components are disjoint and X is compact, [itex]\mathcal C = \{ C_1, \cdots, C_N \}[/itex] must be finite. Define [itex]I = \{1, \cdots, N \}[/itex]. Take for each [itex]n \in I[/itex] a [itex]x_n \in C_n[/itex]. Now define the equivalence relation [itex]\sim[/itex] on X such that [itex]x \sim y \Leftrightarrow \left\{ \begin{array}{l} y \sim x \\ x = y \\ x,y \in \{x_n \}_{n \in I} \end{array} \right.[/itex], or simply put: we identify all the x_n's as one point. This gives us a space Y with the quotient topology and we prove that this is connected.

    Suppose that Y were not connected. Let [itex]Y = A\cup B[/itex] be a separation of Y. Suppose without loss of generality that [itex][x_n] \in A[/itex]. Since A and B are open, we know that [itex]\tilde A = p^{-1}(A)[/itex] and [itex]\tilde B = p^{-1}(B)[/itex] must be open, where p is defined as [itex]p: X \to Y: x \mapsto [x][/itex]. Since [itex]X = \tilde A \cup \tilde B[/itex] and [itex]\tilde A, \tilde B[/itex] are non-empty (due to surjectivity of p) and disjoint (as pre-images), [itex]\tilde A \cup \tilde B[/itex] forms a separation of X. Now since [itex][x_n] \in A[/itex], for all [itex]n \in I: x_n \in \tilde A[/itex], but since each [itex]C_n[/itex] is a component, we have that [itex]n \in I: C_n \subset \tilde A[/itex], giving that [itex]\tilde B = \emptyset[/itex]. Contradiction.

    So even though the formulation of the problem allows a finite number of identifications, one is all you need?
     
  2. jcsd
  3. Jan 7, 2012 #2

    micromass

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    The proof is correct.

    Note that the problem statement said "identification of finitely many points". So, I read that as: take finitely many points and identify them to one point. So you did exactly what the problem statement required of you.
     
  4. Jan 7, 2012 #3
    Ah, that makes sense :) thank you (again)!
     
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