Prove that the Sphere with a Hair in $\mathbb{R} ^{3}$ is not Locally Euclidean at q, hence it can not be a Topological Manifold. 
A fundamental theorem of topology, the theorem on invariance of dimension, states that if two nonempty open sets $U ⊂ \mathbb{R} ^{n}$ and $V ⊂ \mathbb{R} ^{m}$ are homeomorphic, then $n = m$. Prove that the sphere with a hair in $\mathbb{R} ^{3}$ is not locally Euclidean at $q$. Hence it cannot be a topological manifold.

I am new to the theory of manifolds, so I have no idea. 
 A: Suppose the neighborhood $U$ of $q$ is homeomorphic to $B(0,r)\subset\mathbb R^n$ with $q$ mapping to $\bf 0$. Since $U-\{q\}$ has two connected components, then $B-\{0\}$ must be of dimension 1. However since $q$ is on a sphere which is homeomorphic to $\mathbb R^m$, $m>1$, contradiction!
$\square$

http://im0.p.lodz.pl/~kubarski/AnalizaIV/Wyklady/L-Tu-1441973990.pdf
You question is taken from p57 of this book. At page 49 the author proved that a cross is not locally Euclidean at the intersect. 
A: A connected manifold has a unique dimension $n$, and every point of $X$ then has an open neighbourhood homeomorphic to the open unit  ball  $\mathbb D^n\subset \mathbb R^n$.    
However in the pictured $X$ the points different from $q$  on the hair have an open neigbourhood homeomorphic to $\mathbb D^1$ , whereas the points different from $q$  on the sphere have an open neigbourhood homeomorphic to $\mathbb D^2$.    
Since $X$ is connected this proves that it is not a manifold, since it cannot have a unique dimension.
