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!