# Simply connected does not imply contractible. Is there a nice counter example in $R^2$?

The standard counter example to the claim that a simply connected space might be contractible is a sphere $S^n$, with $n > 1$, which is simply connected but not contractible. Suppose that I were interested in a counter example in the plane - does anyone know of a subset of $R^2$ which is simply connected but not contractible?

• Does your definition of contractible require that the resulting point be fixed for all $t$? $\hspace{1.7 in}$ – user57159 Nov 2 '13 at 4:42
• @RickyDemer Does it make a difference? Either definition is fine, just be sure to explain which one you are using. – Lorenzo Najt Nov 2 '13 at 4:43

Consider the topologist sine curve

$$y = \sin \bigg(\frac{1}{x}\bigg),\ 1\geq x>0$$

together with the interval $\{(0, t): |t|\leq 1\}$ and a curve joining this interval with the graph. This is simply connected but is not contractible. You may find the proof of noncontractibility in

http://math.ucr.edu/~res/math205B-2012/polishcircle.pdf

• It's not simply connected - path connectedness is a requirement for simple connectedness. – Lorenzo Najt Nov 2 '13 at 5:09
• Yes, I just made a correction to this. – user99914 Nov 2 '13 at 5:14
• Ah, okay. I see, thanks. – Lorenzo Najt Nov 2 '13 at 5:18
• Is this path connected ? As far as I can recall it is not ... then how is it simply connected ? – user456828 Oct 6 '17 at 6:10
• Topologist sine curve is not. Polish circle is path connected. You couldn't reach the vertical interval from the sine curve in the first case, but you can in the latter, by going in the opposite direction ;) – Ottavio Bartenor Oct 6 '17 at 8:42

John's answer is absolutely correct, but here is an addendum to it: All proper subsets $Z$ of the plane are aspherical in the sense that every continuous map $$f: S^n\to Z, n>1,$$ extends to a continuous map of the ball, $B^{n+1}\to Z$. See the paper One-dimensional sets and planar sets are aspherical by Cannon, Conner and Zastrow.

• This looks interesting, but I don't see how it is an extension of John's answer. (Though I haven't yet read either of the linked papers.) – Lorenzo Najt Nov 2 '13 at 6:38
• @user54092: The point is that the linked paper proves that if such a space Z is simply connected then it is weakly contractible. – Moishe Kohan Nov 2 '13 at 6:42
• Thanks. I didn't know about weak contractibility. – Lorenzo Najt Nov 2 '13 at 6:46
• So the point is that you can't find a counter example in the plane which is a CW complex. That's good to know. – Lorenzo Najt Nov 2 '13 at 6:53