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I need to show that the union of xy-plane and xz-plane, i.e. the set $S:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ or } y=0\rbrace$, is not a surface. Here is my claim,

$\textbf{Claim :}$ Suppose $p$ is the point $(0,0,0)$ and $U:=S\cap B(p,\epsilon)$, where $\epsilon > 0$. Then $U$ cannot be homeomorphic to any open set of $\mathbb{R}^2$.

I need help in proving the above claim. Here are my thoughts: Suppose $U$ is homeomorphic to an open set $V$ of $\mathbb{R}^2$ via homeomorphism $f$.

Now if we remove the $x$-axis from $S$ then it will have $4$ connected components, so if we can show that removal of the image of $x$-axis from $V$ can leave at most three connected components then we are through, is it the case?

Another approach : if we remove the point $p$ from $S$, then the fundamental group of the resulting space is same as the fundamental group of $X:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ and } x^2+y^2=1,\mbox{ or } y=0\mbox{ and } x^2+z^2=1\rbrace$. Can the two fundamental groups $\pi_1(X)$ and $\pi_1(V\setminus \lbrace f^{-1}(p)\rbrace )$ be same? If $V$ were an open ball then I don’t think they are same (though I don’t know the proof), but $V$ is any arbitrary open set.

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How about this: if you remove an arc joining two points on the boundary of a disc, you disconnect the disc; but you can remove an arc joining two points on the boundary of $U$ without disconnecting $U$. – Gerry Myerson Feb 4 '13 at 6:22
@Gerry Myerson: But that would disconnect the disc, it may not disconnnect the open set in which it is sitting and every open set need not be a disc. – pritam Feb 4 '13 at 6:29
Regarding "but V is any arbitrary open set.", it's not arbitrary. Since $V$ is homeomorphic to $S$, it must be connected and simply connected, i.e. it is a ball. Minus a point, it is an annulus. So $\pi_1(V \setminus \{f^{-1}(p)\}) = \mathbb Z$. But $\pi_1(X)$ is something else (try to show this). – Marek Feb 4 '13 at 7:47
Regarding 4 connected components, it's correct. But to prove this, you'd need something on the order of Jordan Curve theorem (to show that number of components outside a simple curve in $\mathbb R^2$ is at most two). – Marek Feb 4 '13 at 7:53
Part of the definition of "surface" is that every point has a neighborhood homeomorphic to a disk. So it's enough to show that $U$ can't be. – Gerry Myerson Feb 4 '13 at 9:24

1 Answer 1

A surface is a space whose points have each an open nghb homeomorphic to an open disc in $\mathbb R^2$. This is equivalent to have an open nghb homeomorphic to an open set in $\mathbb R^2$: the latter always contains an open disc! But the problem is that one cannot say $U=S\cap B$ is homeomorphic to an open disc, only that it contains some open set $W$ of $S$ homeomorphic to a disc. Then $W$ contains another $U'=S\cap B'$ and so on. Thus to argue with fundamental groups $\pi$ one must show that the inclusion $W\subset U$ induces an isomorphism $\pi(W^*)\to\pi(U^*)$ ($*$ means punctured, as usual). Then the first $\pi$ is that of a disc, $\mathbb Z$, and the second one that of the plane minus three points (see below). One needs something here, but Jordan is deeper than these remarks on fundamental groups (my opinion). In fact, to get a contradiction, it is enough to have a surjection $\pi(W^*)\to\pi(U^*)$ and knowing that the second group is not abelian.

Thus, we look at the inclusion $U'\subset U$. Let $\varepsilon$ and $\varepsilon'$ be the radii of $B$ and $B'$, and pick a sphere $E$ centered at $p$ of radius $\frac{1}{2}\varepsilon'$. In this situation the radial retraction of $\mathbb R^3\setminus\{p\}$ onto $E$ provides two deformations retracts $\rho:U^*\to S\cap E$ and $\rho|\!:U'^*\to S\cap E$. In fundamental groups we get isomorphisms $$ \pi(U'^*)\to\pi(S\cap E)\quad\text{and}\quad \pi(U^*)\to\pi(S\cap E), $$ which guarantee the homomorphism $\pi(U'^*)\to\pi(U^*)$ associated to the inclusion $U'^*\subset U^*$ (through which they factorize) is an isomorphism too. But this in turn factorizes through the inclusions $U'^*\subset W\subset U^*$ to give $$ \pi(U'^*)\to\pi(W^*)\to\pi(U^*). $$ Since this composition is bijective, the first arrow must be injective and second arrow onto. This latter fact is the minimum we needed, but remark that playing similarly with another disc $W'\subset U'$ we could obtain that $\pi(U'^*)\to\pi(W^*)$ is onto and then bijective.

Next we have the group $\pi(S\cap E)$, only to give some insight. The space $S\cap E$ is just a couple of circumferences meeting at two points. This is clearly homeomorphic to the union $F$ of an ellipse and a tangent inscribed circumference, sitting in a $\mathbb R^2$. Puncturing three points inside these curves in the obvious way we can deformation retract the plane onto $F$. Thus the homotopy model of $S\cap E$ is a $3$-punctured plane, and one must believe the fundamental group is $\mathbb Z*\mathbb Z*\mathbb Z$, not abelian.

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