[topology] compact, locally connected, quotient topology

[nonequilibrium]

Homework Statement

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.

Homework Equations

The components of a locally connected space are open.

The Attempt at a Solution

And thus the set of components $\mathcal C$ is an open covering of X and since the components are disjoint and X is compact, $\mathcal C = \{ C_1, \cdots, C_N \}$ must be finite. Define $I = \{1, \cdots, N \}$. Take for each $n \in I$ a $x_n \in C_n$. Now define the equivalence relation $\sim$ on X such that $x \sim y \Leftrightarrow \left\{ \begin{array}{l} y \sim x \\ x = y \\ x,y \in \{x_n \}_{n \in I} \end{array} \right.$, 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 $Y = A\cup B$ be a separation of Y. Suppose without loss of generality that $[x_n] \in A$. Since A and B are open, we know that $\tilde A = p^{-1}(A)$ and $\tilde B = p^{-1}(B)$ must be open, where p is defined as $p: X \to Y: x \mapsto [x]$. Since $X = \tilde A \cup \tilde B$ and $\tilde A, \tilde B$ are non-empty (due to surjectivity of p) and disjoint (as pre-images), $\tilde A \cup \tilde B$ forms a separation of X. Now since $[x_n] \in A$, for all $n \in I: x_n \in \tilde A$, but since each $C_n$ is a component, we have that $n \in I: C_n \subset \tilde A$, giving that $\tilde B = \emptyset$. Contradiction.

So even though the formulation of the problem allows a finite number of identifications, one is all you need?

 

[micromass]

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.

 

[nonequilibrium]

Ah, that makes sense :) thank you (again)!