Non orientable phenomena are a consequence about the consideration of the tangent bundles regarding an embedding. One asks if $e:A\to B$ is an embedding then how the tangent bundles $TA$ and $TB$ relate?

For example: we could consider the core (simple close curve) of an cylinder $S^1\times I$ or in a Mobius band $M\ddot{o}$ . First we can observe that if $C_1=S^1\times\{ \frac{1}{2}\}$ has as a regular neighborhood whose boundary is two component disconnected curve (in fact two disjoint circles), while the boundary of a regular neighborhood $N$ of the core curve $C\ddot{o}$ is a single circle: $\partial M\ddot{o}$ .

In terms of tangent bundles we see that we can choose along the cylinder core a consistent normal in the sense that if this curve is traveled then at the end we have the same basis. In contrast with happens in $C\ddot{o}$ which after a full turn we are going to find a reflexion of the normal axe.

Now employing the standard classification of closed surfaces we will construct another kind.

These are the only types of orientable surfaces: $O_0$ the sphere; $O_1$ the two torus; $O_2=O_1\# O_1$ the bitoro; $O_3=O_1\# O_1\# O_1$ the tritoro,... etc, i.e. $$O_g=O_1\# \cdots\# O_1$$

So, with the connected sum device we have:

The projective plane \begin{eqnarray*} {\mathbb{R}}P^2&=&(O_0\setminus{\rm{int}}D)\cup_{\partial}M\ddot{o}\\ &=&D\cup_{\partial}M\ddot{o} \end{eqnarray*}

The Klein bottle \begin{eqnarray*} {\mathbb{R}}P^2\#{\mathbb{R}}P^2 &=&[O_0\setminus( {\rm{int}}D_1\sqcup {\rm{int}}D_2)]\cup_{\partial} [(M\ddot{o})_1\sqcup (M\ddot{o})_2]\\ &=&(M\ddot{o})_1\cup_{\partial}(M\ddot{o})_2 \end{eqnarray*}

If we standarize as $N_1={\mathbb{R}}P^2$ and $N_2={\mathbb{R}}P^2\#{\mathbb{R}}P^2$ , then the genus three non orientable surface is \begin{eqnarray*} N_3&=&{\mathbb{R}}P^2\#{\mathbb{R}}P^2\#{\mathbb{R}}P^2\\ &=&N_2\#{\mathbb{R}}P^2\\ &=&O_1\#{\mathbb{R}}P^2\\ &=&( [O_0\setminus({\rm{int}}D_1\sqcup {\rm{int}}D_2\sqcup {\rm{int}}D_3)]\cup_{\partial} [(M\ddot{o})_1\sqcup (M\ddot{o})_2\sqcup (M\ddot{o})_3]\\ &=&(O_1\setminus{\rm{int}}D)\cup_{\partial}M\ddot{o}\\ &=&(N_2\setminus{\rm{int}}D)\cup_{\partial}M\ddot{o} \end{eqnarray*}

$\bullet\bullet\bullet$