Let $X=\{(x,y)\in \mathbb{R}^2|x^2-y^2-1=0\}$ with the induced euclidean topology. Let's consider the equivalence relation:

$(x,y)\mathscr{R}(x',y') \iff x=\pm x', y'=y$

Let $Y=X /\mathscr{R}$ be the quotient set with the quotient topology $\tau$,

Prove that $(X,\tau)$ connected.

The solution for the connectedness part is given and reads: " Let $\pi$ be the projection on the quotient, $Y=\pi(X \cap \{x>0\})$, then Y is connected". Can someone elaborate o it? I don't get it. I guess they are trying to say that in the quotient it is enought to take one branch of the hyperbole, which is connected and then the projection of a connected set is a connected set, but shouldn't the projection consider the whole hyperbola to make the projection?, since the whole hyperbola is not connected the argument doesn't hold. I feel is not right to take just one branch just because in that way I have a connected space. I know that projecting one branch gives the same projection as projecting the whole hyperbola, but to pass connectedness to the quotient I think you need to take the whole set not part of it.

And what about compactness and Hausdorffnes?, since the hyperbola is not compact I can't say the quotient is compact

Can someone shed some light?

  • $\begingroup$ I' dont understand. Which of the following don't you understand ? $Y = \pi(X\cap x > 0)$, $\pi$ is continuous, $X \cap x > 0$ is connected. If you accept this then Y is the continuous image of a connected set so Y is connected as you wrote. $\endgroup$
    – Digitallis
    Jul 9, 2020 at 22:15
  • $\begingroup$ @Digitallis the first one , which is the the only thing given a solution of the exercise. My guess is that they take the intersection with >0 to make it connected, but I don't feel is right to take part of the set to make the projection, I must take all the set, but if I do I can't use the fact that the projection of a connected set is connected $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 22:21

1 Answer 1


The positive branch of the hyperbola is a connected set, and $Y$ is the image of that set under the continuous map $\pi$, so it must be connected: continuous maps take connected sets to connected sets. What they do to disconnected sets is unpredictable. Sometimes, as is the case here, they take a disconnected set — the whole hyperbola — to a connected set. Sometimes, as in the case of a constant function, they send every disconnected set to a connected set. And sometimes, as in the case of the identity function on any space, they take all disconnected sets to disconnected sets. None of this changes the fact that the image of a connected set under a continuous map is always connected.

For the rest, you can check that the restriction of $\pi$ to the positive branch of the hyperbola is actually a homeomorphism, so $Y$ is Hausdorff but not compact.

  • $\begingroup$ "What they do to disconnected sets...." I guess you meant what they to connect sets $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 22:51
  • $\begingroup$ @J.C.VegaO: No, I meant what I wrote. What they do to connected sets is entirely predictable: they take them to connected sets. $\endgroup$ Jul 9, 2020 at 22:51
  • $\begingroup$ what is the analytical expresion for that restriction? $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 22:52
  • $\begingroup$ @J.C.VegaO: I don’t know what you mean, I’m afraid; what restriction, and what do you mean by analytical expression? $\endgroup$ Jul 9, 2020 at 22:56
  • $\begingroup$ Don't I need a close formula for the homeomorphism?, to prove continuity, injectivity. etc and be able to conclude that it is in fact a homeomorphism $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 22:57

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