A topological group which is also a (not necessarily smooth) manifold is orientable I am trying to show that a topological group which is also a (not necessarily smooth) manifold is automatically orientable. I know of a proof involving transition functions for smooth manifolds, in which case the object in question is a Lie group.
I am using Hatcher's definition of orientability: An $n$-manifold $M$ is orientable if it admits a local orientation $\eta_x$ at each $x\in M$ where $\eta_x$ is a generator of $H_n(M\mid x)\cong \mathbb{Z}$, with the following compatibility property: For each $x\in M$, there is an open ball $x\in B_x\cong \mathbb{R}^n$ so that for every $y\in B_x$, the local orientation $\eta_y$ is the isomorphic image (induced by inclusion of pairs) of the same generator $\eta_{B_x}$ of $H_n(M\mid B_x)$.
I have a clear candidate for such a local orientation, but I am having trouble showing the compatibility: Let $e$ be the identity of the topological group $M$. Choose any generator $\eta_e$ of $H_n(M\mid e)$, and for any $g\in M$, let $\eta_g = L^g_*(\eta_e)\in H_n(M\mid g)$ where $L^g:M\to M$ is left multiplication by $g$ ($L^g$ is a homeomorphism, so it certainly induces an isomorphism on homology).
To start showing the compatibility condition, given $x\in M$, let $B_x$ be any open neighborhood of $x$ homeomorphic to $\mathbb{R}^n$. We are required to show that the following diagram commutes:
$\require{AMScd}$
$$\begin{CD} H_n(M\mid B_x) @>id>\cong> H_n(M\mid B_x) \\ @VV{\cong}V @V{\cong}VV \\ H_n(M\mid x) @>{\cong}>L^{(y^{}x^{-1})}_*> H_n(M\mid y) \end{CD}$$
where the vertical maps are induced by inclusion. Here is where I am stuck. The corresponding diagram on the level of topological spaces certainly does not commute. Any ideas, thoughts, hints, or full solutions are welcome!
 A: The solution by JHF seems correct to me; here is an alternative using the same core idea of paths in the space $M$ yielding homotopies between multiplication maps.
The "top then right" leg of the square is induced by inclusion, which is the same as the map $L^{e}: (M, M - B_{x})\to (M, M - y)$. I will show that this map is homotopic through maps $(M, M-B_{x})\to (M, M-y)$ to $L^{y^{}x^{-1}}$.
The ball $B_{x}\cong \mathbb{R}^{n}$ is path-connected, so we may choose a path $\alpha:I\to M$ from $y$ to $x$ so that the image of $\alpha$ is contained in $B_{x}$. The homotopy $L^{y^{}(\alpha(t))^{-1}}$ is then a homotopy from $L^{e}$ to $L^{y^{}x^{-1}}$, and we must show that every stage of this homotopy is a map of pairs $(M, M-B_{x})\to (M, M-y)$. Suppose therefore that $z\notin B_{x}$. Then if $$y = L^{y^{}(\alpha(t))^{-1}}(z) = y(\alpha(t))^{-1}z,$$ then we must have $$\alpha(t) = z$$ which is impossible since $z\notin B_{x}$ and $\alpha(t)\in B_{x}$.
The "top then right" leg of the square is now simply $L^{y^{}x^{-1}}_{*}$ which certainly causes the square to commute, as desired.
A: It seems like your remaining problem is showing that the degree of the homeomorphism $L^{yx^{-1}}$ is $+1$ (i.e., $L^{yx^{-1}}_*$ is the identity map $\mathbb{Z} \xrightarrow{1} \mathbb{Z}$).  
I claim that the map $L^{yx^{-1}}$ is homotopic to the map $L^e$, which is the identity map $M \to M$.  This is a standard argument.  Since $x$ and $y$ are both contained in the ball $B$ which is path-connected, there exists a path from $x \to y$.  By right-translating by $x^{-1}$, we obtain a path $\gamma: [0,1] \to M$ from $e$ to $yx^{-1}$.  Then $$H(x,t) = L^{\gamma(t)}(x)$$ is a homotopy between $L^e$ and $L^{yx^{-1}}$.  So $\deg L^{yx^{-1}} = \deg L^e = +1$.  
(This question is a duplicate of this question, but I don't think the answer there addresses precisely your concern.  Also, this is also an exercise in Hatcher (3.E.1), which only requires that the topological group has the structure of a finite CW complex.)

EDIT: OK, the fact that you have already defined your local orientations must have escaped my mind when I was writing the above answer.  I still maintain that you're basically done, but maybe the following argument will make you happier?  
Let $g = yx^{-1}$ and consider the commutative diagram of pairs
\begin{array}{ccc}
(M,\emptyset) & \xrightarrow{L^g} & (M, \emptyset) \\
\downarrow & & \downarrow \\
(M, M \setminus \{x\}) & \xrightarrow[L^g]{} & (M, M \setminus \{y\})
\end{array}
In homology this gives us a commuting square:
\begin{array}{ccc}
H^n(M) & \xrightarrow{L^g_*} & H^n(M) \\
\downarrow & & \downarrow \\
H^n(M \mid x) & \xrightarrow[L^g_*]{} & H^n(M \mid y)
\end{array}
I have already argued that $L^g_* = \operatorname{id}$.  Now expand this diagram slightly.
\begin{array}{ccccc}
H^n(M) & & \xrightarrow{\operatorname{id}} & & H^n(M) \\
& \searrow & & \swarrow & \\
\downarrow & & H^n(M \mid B) & & \downarrow \\
& \swarrow & & \searrow & \\
H^n(M \mid x) & & \xrightarrow[L^g_*]{} & & H^n(M \mid y)
\end{array}
We want to show that the bottom triangle commutes.  But all the other triangles commute (the left and right triangles come from maps of spaces, and the top triangle because the top edge is the identity), the outer square commutes, and all the maps are isomorphisms.  So the bottom triangle commutes as well.  
I feel that there must be a simpler way to argue this without all these auxiliary diagrams, but I don't see how right now.  
