Quotient space under equivalence relation

Question

Let $X$ be the quotient space of $\mathbb{R}^2$ received from the following equivalence relation: $$(x_1,x_2) \sim (y_1, y_2) \iff x_1 + x_2 = y_1 + y_2 $$

Now I need to show that $X$ is homeomorphic to $\mathbb{R}$ , so I defined the function $f(x_1, x_2) = x_1 + x_2$ which is obviously bijective and continuous. I just need to prove now that the image under the function of every open set in $\mathbb{R}^2$ is open under $X$.

In order to prove that, I took an open set $U\times V \in \mathbb{R}^2 $ and tried to prove that $f(U \times V)$ is open as well. We get that: $$ f(U \times V) = U+V := \{u+v : u \in U, v \in V\} $$

My question is how to prove that this set is open in X ( if it is ). Thank you!

Answer 1

You want to show that the induced map $X \to \mathbb{R}$ is open, so it's actually enough to show that $f(U)$ is open for open sets $U$ that are saturated, ie closed under the equivalence, ie $b \in U$ for any $a \sim b$ and $a \in U$. But in fact $f : \mathbb{R}^2 \to \mathbb{R}$ is open (with Euclidean topology on both $\mathbb{R}^2$ and $\mathbb{R}$), which is of course sufficient.

It suffices to note that $f(B((x, y), \epsilon))$ contains $B(f(x, y), \epsilon)$, since if $|z - (x + y)| < \epsilon$ then $||(z-y, y) - (x, y)|| < \epsilon$.

Answer 2

Corollary $22.3$ in the topology book by Munkres actually helps visualize the maps and relations. The corollary is as follows:

The proof is almost straightforward. We prove just one direction of the corollary, which we actually need. Let's assume $g$ is a quotient map. We can simply define $f([x])=g(x)$ ($[x]$ is the equivalence class of $x$) and be sure that this map is a well-defined bijection. If $V$ is open in $Z$, since $g^{-1}(V)=p^{-1}(f^{-1}(V))$ and $p$ (the projection map) is a quotient map, we conclude that $f^{-1}(V)$ is open, and hence $f$ is continuous. To show $f$ is a quotient map, we show that $V$ is open in $Z$ if $f^{-1}(V)$ is open in $X^*$. Note that, if $f^{-1}(V)$ is open in $X^*$, the set $p^{-1}(f^{-1}(V))$ is open in $X$ because $p$ is continuous. Since this set equals $g^{-1}(V)$, the latter is open in $X$. Therefore, $V$ must be open in $Z$ because $g$ is a quotient map. So, $f$ is a quotient map, and being a bijection, $f$ is thus a homeomorphism.

Now, we only need to show that $g(x \times y)=x+y$ (similar to what you already defined) from $\mathbb{R}^2$ to $\mathbb{R}$ is a quotient map. We can prove that directly. However, it is not a bad idea to take a look at an exercise of the book:

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To prove the first part of the exercise, let's assume $A=p^{-1}(B)$. We have: $$ f^{-1}(A)=f^{-1}(p^{-1}(B))=(pof)^{-1}(B)=B,$$ which implies if $A=p^{-1}(B)$ is open, then $B$ is open (Be careful not to confuse the usage of the same notation as $f$ and $p$ in the exercise and corollary).

Now, to apply this exercise to $g$, let's define $f$ from $\mathbb{R}$ to $\mathbb{R}^2$ by $f(x)=(x,0),$ which is continuous because $f^{-1} (U\times V)$ is either null or $U$.

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Returning to our main problem, we have proved that $g$ is a quotient map, and by the corollary, $X^{*}$ (the quotient space) is homeomorphic to $Z=\mathbb{R}.$