Vtyjkyuk

Assume that $M$ is an $n$-dimensional smooth manifold. Although I have seen many threads and online handouts dealing with these two different ideas I cannot understand what makes them equivalent. Why the choice of continuous local orientation implies smooth orientation for $M$? The definitions I am using are the following

$mathbf1.)$ Smooth orientation for a manifold $M$ is a maximal atlas $ (U_alpha, phi_alpha) _alpha in A$, such that the transition map $phi_alpha circ phi^-1_beta$, $forall alpha, beta in A$, has positive Jacobian determinant.

$mathbf2.)$ Local (homological) orientation for a smooth manifold $M$ is the choice of generator $[M]_x in H_n(M,M-x) cong mathbbZ$, $forall x in M$. We call such a choice continuous, if for every $x in M$, there exists a neighbourhood $x in U subset M$ and a class $a in H_n(M,M-U)$, such that the inclusion $(M, M-U) to (M,M-x)$, induces a homomorphism $H_n(M,M-x) to H_n(M,M-U)$ which sends $a$ to $[M]_x$. A continuous local orientation for $M$ is called homological orientation.

$2 to 1$

Pick a homological orientation for $M$. Fix a generator of $H_n(Bbb R^n, setminus Bbb R^n setminus 0)$ (equivalently, make sure you know what orientation on $Bbb R^n$ you're using).

For a chart $varphi_alpha: U_alpha to Bbb R^n$ with $varphi_alpha(x) = 0$, note that there is an induced map $(varphi_alpha)_*: H_n(U, U - x) to H_n(Bbb R^n, Bbb R^n setminus 0)$. Because $varphi_alpha$ is a diffeomorphism, $(varphi_alpha)_*$ is an isomorphism. The inclusion map $H_n(U, U - x) to H_n(M, M - x)$ is an isomorphism by the excision theorem.

We say $(U_alpha, varphi_alpha)$ is an oriented chart if the composite $$H_n(M, M - x) xrightarrowi^-1_* H_n(U, U - x) to H_n(Bbb R^n, Bbb R^n setminus 0)$$ sends your chosen generator of the first group to the fixed generator of the last group.

The relevance of the positive Jacobian condition is that a diffeomorphism $f: (Bbb R^n, 0) to (Bbb R^n, 0)$ preserves the generator of $H_n(Bbb R^n, Bbb R^n setminus 0)$ if and only if $det textJac_0(f) > 0.$ In fact, if you have a smooth map $f: Bbb R^n to Bbb R^n$ sending $0$ to $0$, it is homotopic to $textJac_0(f): Bbb R^n to Bbb R^n$, given by setting $f_1-t(x) = fracf(tx)t$ and $f_1(x) = textJac_0(f)(x)$; this is a continuous homotopy of $f$ that sends nonzero points to nonzero points as long as $f$ did the same. (If $f$ did not, restrict to a smaller open set where it did and run the same argument: remember that excision allows us to do this whenever we want.) Now we just need to see the induced maps of linear maps; because $GL_n(Bbb R)$ has exactly two components, determined by their determinant, we just need to check the induced map of two linear maps. The identity induces the identity; a reflection induces $-1$. (Use the long exact sequence of the pair $(D^n, S^n-1)$ and the interpretation of the map on top homology as degree in $S^n-1$.)

Thus the induced map on $H_n(Bbb R^n, Bbb R^n setminus 0)$ is the sign of $det textJac_0 f$. As long as you choose your charts to preserve the homological orientation, your transition maps will always have positive Jacobian determinant.

$1 to 2$

Same recipe as above: this time the rule is that you choose the generator of $H_n(M, M - x)$ using an oriented chart (as opposed to determining which charts are oriented using where the generator goes).