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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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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 parallelizable.

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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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.

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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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 $TM\cong \ell^\ast f^\ast V$. 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 parallelizable.

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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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 parallelizable.

No, this is not true: for each dimension $d \geq 4$, there is an oriented, non-spin manifold 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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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 $\mathbb{T} \to H$ 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, 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 map $\ell: M \to BG$, a bundle isomorphism $TM\cong \ell^\ast f^\ast V$, so that $\ell$ is $2$-connected. This is achieved by surgery below the middle dimension. In particular, $\pi_1 (M)\cong G$, so that $M$ does not have a nontrivial cover, $w_2 (TM) \neq 0$, but the universal cover of $M$ is parallelizable.

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:

1. $H$ is infinite,
2. $H$ does not admit a nontrivial finite quotient,
3. $H$ is acyclic,
4. $H$ 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 $TM\cong \ell^\ast f^\ast V$. 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 parallelizable.