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?
Best Answer
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:
Each is homeomorphic to $(0,1)$.