I am stuck on the following problem which I came across in a recent entrance exam:

In which of the following cases,there is no continuous function $g$ from the set $A$ onto the set $B$ ?

  1. $A=[0,1]\,\,,B=\Bbb R$

  2. $A=(0,1)\,\,,B=\Bbb R$

  3. $A=(0,1)\,\,,B=(0,1]$

  4. $A=\Bbb R\,\,,B=(0,1)$

Can someone explain me how to tackle this?

  • 2
    $\begingroup$ I think just 1...$(0,1)$ and $\mathbb R$ are homeomorphic, so 2 and 4 are out. and for 3 just take a function whose graph is say a 'bump' $\endgroup$ – uncookedfalcon Jun 26 '13 at 3:05


  1. Continuous functions preserve compactness.
  2. Just find a function that s----t---r--e-t-c--h---e----s things.
  3. Try folding $(0,1)$ in half at $\frac12$ and then stretching it a bit.
  4. If you did (2) nicely, it’s a homeomorphism, and its inverse works here.
  • $\begingroup$ @user14111: That’s correct: it is. And one can find examples for (2) and (4) that are not homeomorphisms. But homeomorphisms aren’t too hard to find, and if you find one in either direction, you’ve killed two birds with one stone, as I suggested. $\endgroup$ – Brian M. Scott Jun 26 '13 at 3:25
  • $\begingroup$ @user14111: You’re right: I was thinking homeomorphisms there. Fixed now; thanks. $\endgroup$ – Brian M. Scott Jun 26 '13 at 3:32

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