I was wondering: If $\beta X$ is homeomorphic to $\beta Y$, is it true that $\nu X$ is homeomorphic to $\nu Y$?
Notation: If $f: X\rightarrow \mathbb R$, we denote it's extension by $f^\alpha: \beta X\rightarrow \mathbb R \cup \{\infty\}$
I tried to prove it but I think this won't work: I tried to show that if $\phi$ is the homeomorphism, then the restriction to $\nu X$ will be our homeomorphism. So first, let's try to show that a real point of $\beta X$ is a real point of $\beta Y$: Suppose $f \in C(Y)$. Then $f \circ \phi \in C(X)$. I tried to show that $f^\alpha \circ \phi = (f \circ \phi)^\alpha$ so that would follow easy. In order to do it, I tried to show that these functions restricted to $X$ are the same, but I think I can only say that if $\phi: X\rightarrow Y$. Then I started to think this proposition is false and decided to ask someone.
Is it true?
It's false. Let $X=\mathbb R$, $Y=\beta \mathbb R$. Then $\beta X \approx \beta Y$ but $\nu X \approx \mathbb R$ since $\mathbb R$ is realcompact and $\nu Y\approx \beta \mathbb R$ since $\mathbb \beta \mathbb R$ is compact and therefore realcompact. But $\mathbb R \not \approx \beta \mathbb R$ since the second is compact and the first is not.