A wedge sum of circles without the gluing point is not path connected

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. – Qiaochu Yuan Oct 14 '12 at 8:44
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? – Brian M. Scott Oct 14 '12 at 8:44
@BrianM.Scott The $S^1$ – user42912 Oct 14 '12 at 9:03
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). – Hagen von Eitzen Oct 14 '12 at 9:10
@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? – user42912 Oct 14 '12 at 9:18

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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@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. – Brian M. Scott Oct 14 '12 at 9:31
@user42912: You’re welcome; good luck! – Brian M. Scott Oct 14 '12 at 10:02
@user42912: Yes, that’s exactly what I intended. And those circles are tangent at the origin, so the origin acts as the gluing point. – Brian M. Scott Oct 14 '12 at 10:36
@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. – Brian M. Scott Oct 15 '12 at 8:33
@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.) – Brian M. Scott Oct 15 '12 at 12:47