Hi, everyone,

I resolved the following exercise. I need help reviewing it and giving the correct answer (if mine isn't correct of course).

Let's have:

$\displaystyle I_1=\{(x,y) \in R^2|y=0, 0\leq x\leq 1\}$

$\displaystyle I_2=\{(x,y)\in R^2|x=1, 0\leq y\leq 1\}$

$\displaystyle I_3=\{(x,y)\in R^2|y=x; x,y \in [0 \;1]\}$

$\displaystyle Q=I_1 \cup I_2 \cup I_3$

where the following equivalence relation is defined

$\displaystyle \A (x,y), (x',y') \in Q$, $\displaystyle (x,y)\sim (x',y')$ if $\displaystyle (x,y)=(x',y')$ or $\displaystyle (x, y), (x', y')\in I_1$,

$\displaystyle (x,y),(x',y')\in I_2$

Prove that $\displaystyle \pi : Q\rightarrow Q/\sim$ is closed.

Solution:

$\displaystyle Q$ is connected and compact so even $\displaystyle Q/\sim$ must be connected and compacted. At this point we have to prove that $\displaystyle Q/\sim$ is $\displaystyle T_2$ (Hausdorff). That is to prove that the set

$\displaystyle F= \{ (x;y),(u;v): \pi(x;y)=\pi(u;v) \}=\{ (x;y),(u;v): (x;y)\sim(u;v) \}$ is closed on $\displaystyle Q\times Q$.

$\displaystyle C_1=\{ (x;y),(x;y)\in Q\times Q| (x;y)\in Q}$, $\displaystyle C_2=\{ (x;0),(x;0)\in Q\times Q| x\in[0\;1]\}$

$\displaystyle C_3=\{ (1;y),(1;y)\in Q\times Q| y\in[0\;1]\}$.

$\displaystyle C_1$ is closed because $\displaystyle Q$ is $\displaystyle T_2$. $\displaystyle C_2$, $\displaystyle C_3$ are closed being subspaces of space $\displaystyle T_2$ and immages of the compact $\displaystyle [0\;1]$ on $\displaystyle R$ through continuous applications. Because $\displaystyle F=C_1 \cup C_2 \cup C_3 $ then even $\displaystyle F$ will be closed, this means that $\displaystyle Q/\sim$ is $\displaystyle T_2$. I conclude that $\displaystyle \pi$ is closed.