Non-orientable one dimensional manifold. I was trying to solve a question from Hatcher's book in section 3.3.
Question is:
Show that there exist a non-orientable 1-dimensional manifold if Hausdroff condition is droped from the definition of manifold.
I know real line with two origin is a 1-dimensional non-Hausdroff manifold.
But I don't know how to show that this is orientable or not?
I will apprecite if somebody explain this to me using the local homology definition of orientability.
 A: You can make a "branch", the real line with doubled non-negative side. You can then have the two branches meet, and you have a "loop on a string". This is non-orientable. More specifically, we have the disjoint union of two copies of $(\infty, 1]$, divided by the relation $\sim$ given by $a\sim b$ iff both of the following are satisfied:


*

*$a$ and $b$ are in different components

*Either of the following is satisfied:


*

*$a$ and $b$ are negative and $a = b$

*$a$ and $b$ are non-negative and $a + b = 1$.



This gives something that looks like this:

In the case of $1$-dimensional manifolds, a choice of orientation is a choice of "backwards" and "forwards". In the relative homology group $H_1(M, M\setminus \{x\})\approx \Bbb Z$, for any $1$-manifold $M$ and any point $x \in M$, elements are represented by $1$-chains in $M$, and the homology counts how many times the chain passes through $x$, with direction. Therefore, a generator of the local homology group corresponds to deciding which direction through $x$ should be the positive one.
With that in mind, here is a homological proof that the above manifold $M$ is non-orientable. First, some convensions. I will use superscripts $^+$ and $^-$ for the two different branches of the non-negative real line. I will use the notation $(a, b)$ for the singular $1$-chain in $M$ with starting point $a$ and endpoint $b$, parametrized linearly. Note that we do not necessarily have $a < b$, and also that one or both of $a$ and $b$ might have a superscript.
Take the $1$-chain $(-1, -0.3)$. It represents a generator for $H_1(M, M\setminus \{-0.5\})$ and therefore gives an orientation at $-0.5$. The $1$-chain $(-1, 0.7^+)$ represents the same generator, but this chain also represents a generator of $H_1(M, \{0.5^+\})$, and thus gives us compatible orientations at the two points $-0.5$ and $0.5^+$.
The $1$-chain $(0.3^+, 0.7^+)$ represents the same element in $H_1(M, \{0.5^+\})$, and therefore gives the same orientation. Translating to the other branch, we have that $(0.7^-, 0.3^-)$ represents a generator of $H_1(M, M\setminus \{0.5^-\})$. This is just a translation, so we still have the same orientation.
In $H_1(M, M\setminus \{0.5^-\})$, the chain $(0.7^-, 0.3^-)$ represents the same generator as $(0.7^-, -1)$. Here comes the non-orientability: This chain represents the same generator of $H_1(M, M\setminus\{-0.5\})$ as does $(-0.3, -1)$, which is the negative of the chain we started with, $(-1, -0.3)$.
To recap, we chose a $1$-chain $(-1, 0.7^+)$ that generates the local homology at $-0.5$, and see which generator that chain represents at $0.5^+ = 0.5^-$. We then took another $1$-chain $(0.7^-, -1)$ that represents the same generator at $0.5^\pm$, but another generator at $-0.5$ (we used a few intermediate chains, but they're not really necessary). Therefore there is no consistent choice of generators of the local homology groups, and the manifold is non-orientable.
