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Connected proof.

Siron

Active member
Jan 28, 2012
150
Hi, can someone help me with the following question(s)?
1. Let $p:X \to Y$ be a quotient map. Suppose that [tex]p^{-1}(y)[/tex] is connected for each [tex]y \in Y[/tex]. Show [tex]X[/tex] is connected if an only if [tex]Y[/tex] is connected.

2. Let [tex](X,\mathcal{T})[/tex] be a topological space and $A,B \subseteq X$. Suppose $A$ is open, $\mbox{cl}(A) \subseteq B$, $B$ is connected and $\partial A$ is connected. Proof $B\setminus A$ is connected.

The problem is I don't know a good idea how to prove a set is connected. For the first question, it's clear that $Y$ is connected when $X$ is connected because $p$ is a quotient map. But I don't know how to prove the other implication.

I have no clue for the second question.

Anyone?
 

hmmm16

Member
Feb 25, 2012
31
For the first part I think that we can have the following:

Suppose that $Y$ is connected and $X$ is not connected. Then $X=U_1\cup U_2$ for some open $U_1,U_2\subset X$ and $U_1\cap U_2=\emptyset$

Now note that as $p^{-1}(y)$ is connected then we have that $p^{-1}(y)\subset U_1$ or $p^{-1}(y)\subset U_2$

Now define $V_1=\{y\in Y|p^{-1}(y)\subset U_1\}$ and $V_2=\{y\in Y|p^{-1}(y)\subset U_2\}$

Then we have that $V_1\cup V_2=Y$ and that $V_1\cap V_2=\emptyset$

So all we need to show is that $V_1,V_2$ are open and we have a contradiction.


Now we have that $p^{-1}(V_1)=U_1$ and $p^{-1}(V_2)=U_2$b which are both open in $X$ then from the defintion of a quotient map we have that:

$V\subset Y$ is open iff $p^{-1}(V)$ is open in $X$. So we have a contradiction and we are done.

However I am afraid that for the second one I am pretty stumped.