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 topological group which is also a (not necessarily smooth) manifold is orientable
In fact beside your question there is a beautiful theorem in homotopy theory. This theorem due to T. Bauer, N. Kitchloo, D. Notbohm and E. K. Pedersen guarantees that any loop space $X=\Omega B$ where $B$ is a $CW$-complex and such that $H\_\*(X)=\oplus H\_i(X;\mathbb{Z})$ is a finitely generated abelian group is homotopy equivalent to a compact, smooth, parallelizable manifold ("Finite loop spaces are manifolds" Acta Math. 2004).


This theorem applies to your case where $X=G\simeq \Omega BG$. 


The very first step in the proof of this result which is very closely related to your question is to prove that $X$, thus $G$ in your case, is a Poincaré duality space. The proof is completely algebraic and uses the fact that for any field $\mathbb{F}$ the cohomology algebra is a finitely generated, connected graded Hopf algebra. Then you use Borel's classification that tells you that $H^\*(G;\mathbb{F})$ is a tensor product of exterior algebras and truncated polynomials algebras and get a family of top classes $[G]\_p\in H\_{n\_p}(G;\mathbb{Z}/p\mathbb{Z})$ for an integer $n\_p$. Then you show that all these $n\_p$ are equal to a fixed integer $n$ and that $[G]\_p$ is the reduction $mod(p)$ of an integral fundamental class whose cap product induces a Poincaré duality isomorphism. All details and references are given in the introduction of Bauer, Kitchloo, Notbohm, Pedersen's paper. 


Edit: It is also certainly worth recording Gleason, Montgomery-Zippin theorem, namely:


"A topological group G underlies a (unique) Lie group structure if and only if the underlying space of G is a topological manifold." 


