A cross in $\mathbb R^2$ is not a topological manifold

Question

In Loring Tu's Introduction to Manifolds pg. 49, it was said that the cross in $\mathbb R^2$

is not a topological manifold because it is not locally Euclidean at the intersection $p$.

In the proof, it was written:

Suppose the cross is locally Euclidean of dimension $n$ at the point $p$. Then $p$ has a neighborhood $U$ homeomorphic to an open ball $B :=B(o,\epsilon) \subset\mathbb R^n$ with $p $ mapping to $0$.

But the definition of locally Euclidean is given on pg. 48 as:

A topological space $M$ is locally Euclidean of dimension $n$ if every point $p$ in $M$ has a neighborhood $U$ such that there is homeomorphism $\phi $ from $U$ onto an open subset of $\mathbb R^n$.

The definition suggests that there may be only one neighbohood $U$ containing $p$ that can be mapped onto an open subset of $\mathbb R^n$, which need not be an open ball. Why can the author assume that for the cross, there can be multiple neighborhoods $U$ containing $p$ that each maps onto an open ball of arbitrary radius $\epsilon?$

Answer 1

Locally Euclidean implies more than you think: let $p \in M$ and let $U$ be the promised neighbourhood of $p$ in the definition.

So there is a homeomorphism $h: U \to O$ where $O \subseteq \Bbb R^n$ is (Euclidean) open.

So in particular $h(p) \in O$ and so for some $r>0$ we have that $B(h(p), r) \subseteq O$ by the definition of the Euclidean topology. But if $V$ now is an open ball around $h(p)$ of radius $0 < r' \le r$ we have that

$V':=h^{-1}[V]$ is open in $U$ and $h\restriction_{V'}: V' \to V$ is also a homeomorphism (simple check using that the restriction of an open map to open sets is still open, restrictions of continuous maps remain continuous etc. All basic facts).

So we can pick the codomain set to be some open ball of small enough radius if that is convenient in the proof. This proof is an example where it's used.

Answer 2

There is an open ball around $\phi(p)$ contained in $\phi(U)$, since $\phi(U)$ is open. The preimage of this ball is an open neighborhood of $p$ homeomorphic to the ball, where $\phi$ restricted to said preimage is the homeomorphism. You might want to verify a few things yourself, such as:

1. The image of the preimage is again the open ball.
2. The restricted map is injective.
3. The restricted map is continuous and open.
4. Thus it is a homeomorphism onto its image (the ball)