If the universal cover of a manifold is spin, must it admit a finite cover which is spin?
No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.
The reason is simply that there are finitely presented groups which have no nontrivial finite quotient. One example is Higman's group $H$, see https://en.wikipedia.org/wiki/Higman_group.
The key features of $H$ are:
- $H$ is infinite,
- $H$ does not admit a nontrivial finite quotient,
- $H$ is acyclic,
- $H$ has a classifying space $BH$ which is a finite $2$-dimensional CW-complex.
The proof of 1,2 can be found in Tao's blog https://terrytao.wordpress.com/2008/10/06/finite-subsets-of-groups-with-no-finite-models/, and the proof of 3,4 in ''The topology of discrete groups'' by Baumslag, Dyer, Heller).
Now pick an element $1 \neq x \in H$, which induces an injective homomorphism $\mathbb{Z} \to H$ and form the amalgamated product $G=H \ast_{\mathbb{Z}} H$. The group $G$ is infinite and has no nontrivial homomorphism to a finite group $F$, since any homomorphism $G \to F$ must vanish on the two copies of $H$.
The pushout $BH \cup_{S^1} BH$ is aspherical by Whiteheads asphericity theorem and hence a model for $BG$. I have designed things so that $H_2(BG) \cong \mathbb{Z}$ and all other homology groups are trivial. In particular, $G$ is perfect, and the Quillen plus construction $BG^+$ must be homotopy equivalent to $S^2$, so that there is a homology equivalence $f:BG \to S^2$. Now let $V \to S^2$ be the nontrivial oriented vector bundle of rank $d$, which has $w_2 (V) \neq 0$. It follows that the vector bundle $f^\ast V \to BG$ is not spin. $BG$ has no nontrivial cover, and $BG$ is aspherical, so the pullback of $f^\ast V$ to the universal cover is trivial.
Now there exists, when $d \geq 4$, a closed $d$-manifold $M$ with a $2$-connected map $\ell: M \to BG$ and a bundle isomorphism $TM\oplus \mathbb{R}\cong \ell^\ast f^\ast V \oplus \mathbb{R}$. This is achieved by surgery below the middle dimension. In particular, $\pi_1 (M)\cong G$. Hence $\pi_1(M)$ has no nontrivial finite index normal subgroup, and therefore no nontrivial finite index subgroup at all. It follows that $M$ does not have a nontrivial finite cover.
By construction, $w_2 (TM) \neq 0$, but the universal cover of $M$ is stably parallelizable.
As promised, here is my solution based on the Davis trick. First, there is a very general construction of PL aspherical 4-manifolds (it also works in higher dimensions). Start with a finite aspherical 2-dimensional CW complex $W$. Up to homotopy, $W$ always embeds in the Euclidean 4-space $E^4$ (I think, this is due to Stallings). Take such an embedding and let $N=N(W)$ denote a regular neighborhood of $W$ in $E^4$. Now, apply "Davis trick" to $N$: Introduce a reflection orbifold structure on the boundary of $N$ such that the corresponding stratification of the boundary is dual to a triangulation of $\partial N$. The resulting orbifold ${\mathcal O}$ is very good (admits a finite orientable manifold-covering $M\to {\mathcal O}$) and its universal covering (same for $M$ and for ${\mathcal O}$) is contractible. As a bonus, $\pi_1(W)$ embeds in $\pi_1(M)< \pi_1({\mathcal O})$. For details see
Mess, Geoffrey, Examples of Poincaré duality groups, Proc. Am. Math. Soc. 110, No. 4, 1145-1146 (1990). ZBL0709.57025.
and, of course, the original paper by Mike Davis from 1983. (Actually, it was Bill Thurston who came up with this trick in the context of 3-manifolds: He used it for the proof of his hyperbolization theorem.) This construction allows one to embed 2-dimensional finitely presented groups with "exotic properties" in fundamental groups of closed aspherical PL manifolds.
I will use a relative version of this construction. Start with a closed connected oriented surface of genus $\ge 1$; I'll take the torus $T^2$. Let $E\to T^2$ be the 2-disk bundle over $T^2$ with the Euler number $\pm 1$. The boundary of the 4-manifold $E$ is a 3-dimensional Nil-manifold: The total space of a nontrivial circle bundle over the torus. The group $\pi_1(\partial E)$ has two generators $a, b$, and $\pi_1(\partial E)$ has the presentation $$ \langle a, b| [a,b]=t, [a,t]=1, [b,t]=1\rangle. $$ Represent $a, b$ by simple disjoint loops $\alpha, \beta$ in $\partial E$. Now, take your favorite finite 2-dimensional aspherical complex $W$ whose fundamental group is nontrivial and has no proper finite index subgroups (I care only about the homotopy type of $W$). The standard example is the presentation complex of Higman group. But there are many other examples. As before, embed $W$ in $E^4$, take a regular neighborhood $N$ of $W$ in $E^4$. Then $\pi_1(\partial N)$ maps nontrivially to $\pi_1(W)$. Pick two simple loops $\alpha', \beta'\subset \partial N$ which map nontrivially to $\pi_1(W)$ (you can take the same loop).
Now, take two copies $N_a, N_b$ of $N$ and attach them to $E$ by identifying a regular neighborhood of $\alpha'$ to that of $\alpha$ for $N_a$ and identifying a regular neighborhood of $\beta'$ to that of $\beta$ for $N_b$. The result is a compact PL aspherical 4-manifold with boundary $Z$. The $\pi_1(Z)$ is an amalgam of $\pi_1(E)\cong {\mathbb Z}^2$ with two copies of $\pi_1(N)$ (along infinite cyclic subgroups). For each homomorphism to a finite group $$ \phi: \pi_1(Z)\to \Phi $$ the subgroups $\pi_1(N_a), \pi_1(N_b)$ will have to map trivially. Hence, $a$ and $b$ will have to map trivially as well. Since $a, b$ generate $\pi_1(E)$, $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Now, apply Davis trick to $Z$. The result is an orbifold ${\mathcal O}$. Since $Z$ was aspherical, so is ${\mathcal O}$ (i.e. it has contractible universal covering space).
Take a finite orientable manifold-covering $M\to {\mathcal O}$. Then $M$, of course, has contractible (hence, spin) universal covering. I claim that $M$ has no finite spin-covering spaces. Indeed, for each finite-sheeted covering $p: M'\to M$, the manifold $int(Z)\subset M$ has to lift trivially; more precisely, $p$ restricts to a trivial covering $$ p^{-1}(int Z)\to int Z.$$ This is because $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Thus, $M'$ contains a copy of $E$. In particular, $M'$ contains a 2-torus with odd self-intersection, i.e. the intersection form of $M'$ is not even, i.e. $M'$ is not spin.
I was working in the PL category but in dimension 4, PL is the same DIFF, so you get a smooth example as well.
Edit. Lemma. Let $M$ be a triangulated manifold, $W\subset M$ is a subcomplex and $N=N(W)$ is the regular neighborhood of $W$ in $M$. Then the inclusion map $W\to N$ is a homotopy-equivalence; if $W$ is connected and has codimension $\ge 2$ in $M$ then $\partial N$ is connected and the induced map $\pi_1(\partial N)\to \pi_1(W)$ is surjective.
Proof. The homotopy-equivalence part is standard and holds for general simplicial complexes $M$, not just for manifolds. Moreover, the inclusion map $\partial N\to (N \setminus W)$ is also a homotopy-equivalence. (Both are proven using "straight-line homotopy.")
I will prove the second part. Take an arc $\alpha$ in $N$ connecting two points $x, y\in \partial N$. Since $W$ has codimension $\ge 2$, taking $\alpha$ in general position, we see that it will be disjoint from $W$, hence, is homotopic relative to $\{x, y\}$ to an arc in $\partial N$. (I am using here and below the h.e. $\partial N\to N-W$.) Thus, $\partial N$ is connected. Next, let $\alpha$ be a loop in $N$ based at $x\in \partial N$. By the same argument, $\alpha$ is homotopic to a loop based at $x$ and contained in $N-W$, hence, to a loop in $\partial N$.