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Note: for this question a topological manifold is defined to be a locally Euclidean, second countable, Hausdorff space. Also, I am using the subscript $x$ and $y$ just to keep track of which copy of $\mathbb{R}$ I am calling the $x$ and $y$ axis, respectively.

Intuitively I can see why the wedge sum of the $x$ and $y$ axes are not a manifold, since they do not have a well defined dimension. I am having trouble making this idea more rigorous.

Let $\mathbb{R}_x$ denote the $x$-axis and $\mathbb{R}_y$ denote the $y$-axis. Also, let $0_x$ and $0_y$ be the zero on the $x$ and $y$ axis, respectively. The wedge sum of the $x$ and $y$ axes is defined as

$$\mathbb{R}_x\vee\mathbb{R}_y=\mathbb{R}_x\amalg\mathbb{R}_y/(0_x \sim 0_y).$$

Any open neighborhood of $0$ in $\mathbb{R}_x\vee\mathbb{R}_y$ would have to be open when intersected with both $\mathbb{R}_x$ and $\mathbb{R}_y$. This leads me to believe that the open neighborhoods of $0$ in $\mathbb{R}_x\vee\mathbb{R}_y$ are wedge sums of intervals containing $0_x$ and $0_y$ in the $x$ and $y$ axes. Then is $\mathbb{R}_x\vee\mathbb{R}_y$ not a topological manifold because it is not locally Euclidean around $0$?

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1 Answer 1

up vote 5 down vote accepted

Yes, that's correct. To finish the proof, you need to show that a wedge sum of two open intervals is not homeomorphic to an open disc in $\mathbb{R}^n$ for any $n$. This can be done, for example, by considering what happens if you delete the origin from the wedge sum: the resulting space will have four connected components. Deleting a point from an open disc in $\mathbb{R}^n$ will leave nothing (if $n=0$), two components (if $n=1$), or one component (if $n\ge 2$), but never four components.

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