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I am trying to prove that a wedge sum of two circles is not a topological manifold, to do so I am showing that the wedge sum without the gluing point is not path connected while the $\mathbb R^2$ without a point is path connected, this implies that the wedge sum is not a manifold since it is not homeomorphic to $\mathbb R^2$.

However, my question is how to prove formally that a wedge sum of circles without the gluing point is not path connected, I can see that intuitively, but it seems very difficult to prove formally. I've tried so hard, but I do not know even how to begin to solve the problem. I think if I can solve this question, I can use it as model to solve the similar ones, Anyone can help me please?

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    This doesn't show that the wedge sum is not a manifold.2012-10-14
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    When you say *circle*, do you mean $S^1$ (which is a circle), or do you mean a closed ball in $\Bbb R^2$, which is actually a disk?2012-10-14
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    @BrianM.Scott The $S^1$2012-10-14
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    The neighbourhood of $S^1\lor S^1$ around any point other than the glueing point looks like $\mathbb R^1$, so it is of no use to show that no neighbourhood of the glueing point loooks like $\mathbb R^2$. But you can still tke this path: Every neighbourhood of a point in $\mathbb R^1$ has a subneighbourhood such that it deleting one point produces exactly two connected components (not four).2012-10-14
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    @HagenvonEitzen yes, I understood, thank you, your comment clarifies me a lot. However my problem remains unsolvable for me, how can I prove formally that the wedge sum of the two cicles has 4 components?2012-10-14
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    @user42912: You can't, since the wedge sum of the two circles has one component.2012-10-14

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The wedge sum of two copies of $S^1$ is not homeomorphic to $\Bbb R^2$ with or without the gluing point. With the gluing point it has a point, the gluing point, whose removal disconnects it, while $\Bbb R^2$ has no such point. Without it it’s disconnected, while $\Bbb R^2$ is connected. In any case, as Qiaochu said, this doesn’t show that the wedge sum of two circles is not a manifold.

HINT: Does the gluing point have a neighborhood homeomorphic to $R^n$ for any $n$? Note that if $p$ is the gluing point, $p$ has arbitrarily small neighborhoods that look like $+$; is this true of any point in any $\Bbb R^n$? Think about connectedness and numbers of connected components.

Added: We can realize $S^1\lor S^1$ as $$X=\Big\{\langle x,y\rangle\in\Bbb R^2:(x+1)^2+y^2=1\text{ or }(x-1)^2+y^2=1\Big\}\;,$$ with $p=\langle 0,0\rangle$ as the gluing point. (You shouldn’t have too much trouble writing down a homeomorphism between the formal description of $S^1\lor S^1$ and this $X$.) A typical small open neighborhood of $p$ in $X$ is $N_\epsilon=X\cap B(p,\epsilon)$, where $\epsilon>0$, and $B(p,\epsilon)=\{q\in\Bbb R^2:\|q\|<\epsilon\}$, the open $\epsilon$-ball centred at $p$. If $\epsilon\le2$, $N_\epsilon\setminus\{p\}$ has the following four components:

  • $\{\langle x,y\rangle:(x-1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x,y>0\}$
  • $\{\langle x,y\rangle:(x-1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x>0\text{ and }y<0\}$
  • $\{\langle x,y\rangle:(x+1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x<0\text{ and }y>0\}$
  • $\{\langle x,y\rangle:(x+1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x,y<0\}$

Each is homeomorphic to $(0,1)$.

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    yes the $\mathbb R^n$ minus a point is always connected if $n\gt 1$ while the wedge sum of the two cicles minus the gluing point has 4 components. However my problem is how to prove formally that the number of components of the wedge sum minus the gluing point of two cicles is 4.2012-10-14
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    @user42912: It could actually be $2,3$, or $4$, depending on what nbhd of $p$ you look at. $(S^1\lor S^1)\setminus\{p\}$ has two components; removing $p$ from a nbhd that includes all of one $S^1$ and an arc of the other leaves three components. Give me a few minutes, and I’ll add something about this to my answer.2012-10-14
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    thank you, I will work hard now to prove these statements.2012-10-14
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    @user42912: You’re welcome; good luck!2012-10-14
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    in order to prove that $S^1\lor S^1$ is homeomorphic to X, I realized I need the abstract concept of $S^1$ in this context. Can you help me?2012-10-14
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    Can I defined $S^1$ as one of the cicles you wrote such as $(x+1)^2+y^2=1$ or $(x-1)^2+y^2=1$?2012-10-14
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    @user42912: Yes, that’s exactly what I intended. And those circles are tangent at the origin, so the origin acts as the gluing point.2012-10-14
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    I'm trying so hard to prove this homeomorphism, but it seems very hard. I've noticed that we need just to prove the homeomorphism between any open subset of the circle and any open interval of $\mathbb R$, because I think we can find easily the homeomorphism between (0,1) and this open interval. So my first thought was the map $t\to (cost,sint)$ but it failed, so do you have any hint to help me to prove it? it seems hard to find.2012-10-15
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    @user42912: It doesn’t fail: $t\mapsto\langle\cos 2\pi t,\sin 2\pi t\rangle$ maps $(0,1)$ homeomorphically onto all of the unit circle except the point $\langle 1,0\rangle$, so by choosing the right open subinterval of $(0,1)$, you can get any open arc of the unit circle.2012-10-15
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    This map is continuous because the components are continuous, I'm trying to prove that this function is open. When I take a open interval U in the domain, how can I prove that f(U) is a open subset in S minus (0,1)?,in another words, how do I prove that f(u) is the intersection of a open ball and the circle? it looks crazy to prove that.2012-10-15
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    @user42912: That’s the hard way. Just show that its inverse is continuous; the inverse maps the point with polar coordinates $\langle 1,\theta\rangle$ to $\frac{\theta}{2\pi}\in(0,1)$, and that’s clearly a continuous map. (I’m about to go to bed, so it’ll be a few hours before I’m back.)2012-10-15
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    ok, I understood everything, I have now only one problem, did you proved for a neighborhood of p, the problem is we have to prove to any neighborhood of p, where's my mistake?2012-10-16
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    @user42912: Every open nbhd of $p$ contains an open nbhd of the form $+$, one that falls into $4$ components when you remove $p$. No point of $\Bbb R$ has even one such nbhd. Therefore no open nbhd of $p$ is homeomorphic to $\Bbb R$.2012-10-16
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    and about $\mathbb R^n$ for $n\gt 1$? we know that $\mathbb R^n$ minus a point is connected, but is it possible to have opens subsets in $\mathbb R^n$ when we remove a point it has four components?2012-10-16
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    @user42912: If $n>1$, and you remove two points from $\Bbb R^n$, what’s left is connected. If $V$ is any connected open nbhd of $p$, you can always find two points of $V$ whose removal disconnects $V$. Therefore $V$ cannot be homeomorphic to $\Bbb R^n$. And if $V$ isn’t connected, then it certainly isn’t homeomorphic to $\Bbb R^n$. Thus, $p$ has no open nbhd homeomorphic to $\Bbb R^n$ for $n>1$.2012-10-16
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    I think any point we remove from V disconnected V, no?2012-10-16
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    @user42912: If $V$ includes all of one of the circles, you can remove a point on that circle (other than $p$ itself) without disconnecting $V$. That’s why I used two points.2012-10-16
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    so and we have to remove these two point from the same circle to disconnect the space.2012-10-16