1
$\begingroup$

Let $X,Y$ be topological spaces, $\tilde~$ equivalence relation on space $X$ and $\tilde~'$ equivalence relation on space $Y$.

If $f:X\rightarrow Y$ is homeomorphism, where $x_{1}\tilde~ x_{2}\Leftrightarrow f(x_{1})\tilde~' f(x_{2})$, so $f$ induces a homeomorphism $X/\tilde~\rightarrow Y/\tilde~'$.

I am using the theorem that says that bijection $f:X\rightarrow Y$ is a homeomorphism, iff $U\subseteq X$ open iff $f(U)\subseteq Y$ open.

Now, that part where I must show the equivalence mentioned above is clear.

But my problem is the bijectivity of $\overline{f}:X/\tilde~\rightarrow Y/\tilde~'$.

The condition between the equivalence relations means that for every point in any equivalence class it's image is in the image of it's equivalence class. That and the fact $f$ is homeomorphism should give me enough information to say that $\overline{f}$ is indeed bijection. But I am not sure.

Thoughts, ideas?

$\endgroup$
4
$\begingroup$

$\def\si{\mathord\sim}$To see that $\bar f$ is one-to-one, suppose $\bar f(x_1/\si) = \bar f(x_2/\si)$, that is $f(x_1) = f(x_2)$ by definition of $\bar f$, therefore $f(x_1) \sim' f(x_2)$, hence $x_1 \sim x_2$. So we have $x_1/\si = x_2/\si$.

To see that $\bar f$ is onto, suppose $y/\si' \in Y/\si'$, as $f$ is onto, there is $x \in X$ with $f(x) = y$, but then $\bar f(x/\si) = y/\si'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.