What are some examples of $4$-manifolds $M$ for which the class $w_3(TM)\in H^3(M;\mathbb{Z}/2)$ is nontrivial? Is there a mapping torus with this property?
Motivation: I am wondering whether any such $4$-manifolds can be "built out of" a $3$-manifold by the mapping torus construction, despite the fact that $w_3$ vanishes on $3$-manifolds. In asking myself this, I realized my that go-to examples of $4$-manifolds -- $\mathbb{R} P^4$, $\mathbb{C} P^2$, and $K3$ -- all have trivial $w_3$.
On
Here is an idea I came up but I was only able essentially reduce it to this question. I had hoped to prove that the mapping torus of a diffeomorphism of an orientable $3$-manifold is zero using the fact that orientable $3$-manifolds have trivial tangent bundles and the definition of Stiefel-Whitney classes as obstruction classes. In the end this gave a good method to produce counterexamples I think.
Let $M$ denote an orientable $3$-manifold and $f$ and orientation preserving diffeomorphism. I will denote the mapping Torus by $T_f$ and the inclusion of some fiber by $\iota\colon M\to T_f$. Fix a trivialization of $T\iota(M)$, which is possible by the aforementioned fact. Now we know that $\iota^*(w_i(S_f))=w_i(\iota^*TS_f)=w_i(TM\oplus \mathbb{R})=0$ we conclude that $w_i(S_f)$ comes from some class in $H^i(S_f,\iota(M);\mathbb{Z}/2\mathbb{Z})$. Using excision the inclusion of the pair $(M\times I,M \times \partial I)\to (S_f,\iota(M))$ induces an isomorphism on cohomology. Therefore we have to understand how the Stiefel-Whitney classes of $(M\times I,\partial M\times I)$ behave.
Since $T(M\times I)\cong \pi^* TM\oplus \mathbb{R}$, where $\pi$ denotes the projection $M\times I\to M$ and this splitting respects the fixed framing at $M\times \partial I$, we have to understand $w_3(\pi^*TM,\pi^*TM|_{M\times \partial I})$. Note that this is the mod $2$ reduction of the relative Euler class. Furthermore note that if we fix some non-vanishing section $\phi$ of $\pi^*TM|_{M\times \{0\}}$ then the section at $\pi^*TM|_{M\times \{1\}}$ is given by $f_* \phi((x,1))=Df_{f^{-1}(x)}(\phi(f^{-1}(x))$. All in all this should imply that $w_3$ is the mod $2$ reduction of the obstruction class for a homotopy between $\phi$ and $f_*(\phi)$.
Therefore we are left with the question how do homotopy classes of non-vanishing vector fields on orientable $3$-manifolds behave under diffeomorphisms of said manifold, which is exactly the aforementioned question. Nevertheless note that $[M,S^2]$, which is the set of homotopy classes of vector fields, surjects quite naturally to $H^2(M;\mathbb{Z})$. So maybe it is possible to deduce the existence of a vector field $\phi$ and a diffeomorphism $f$ such that the obstruction class for a homtopy between $f_* \phi$ and $\phi$ is non-zero mod $2$ using the action of $f$ on $H^2(M)$, but I'm tired right now so I will think about this last part tomorrow.
This is a partial answer, showing what one should demand if they were to try to find an example with $w_3(T_f) \neq 0$: If $M$ is an orientable 4-manifold then $w_3(M) = 0$. So $f$ would have to either be an orientation-reversing diffeomorphism or a diffeomorphism of a non-orientable manifold.
To prove this, use Wu classes. These are the classes $\nu_i \in H^i(M;\Bbb Z/2)$ for which the two maps $H^{n-i}(X;\Bbb Z/2) \to \Bbb Z/2$ given by $\nu_i \cdot x$ and $\text{Sq}^i x$ agree. Wu's theorem is that we have the property $$\sum_{i=0}^{\lfloor k/2\rfloor} \text{Sq}^{k-i} \nu_i = w_i.$$
We see from the definition that $\nu_3 = 0$ because $\text{Sq}^3$ vanishes on classes of degree less than $3$, and we see from orientability that $\nu_1 = 0$, and hence $\nu_2 = w_2$. Therefore we have $w_3 = \text{Sq}^1 w_2$. (In fact, this is true for an arbitrary 4-manifold; one needs to argue that $(\text{Sq}^1)^2 w_1$, which in principle contributes, is always zero). The operation $\text{Sq}^1$ is sometimes better known as the Bockstein map. This map factors as the composite of the integral Bockstein $\beta_{\Bbb Z}: H^2(M;\Bbb Z/2) \to H^3(M;\Bbb Z)$ and reducing coefficients modulo 2, so it suffices to show that if $M$ is an oriented closed 4-manifold, we have $\beta_{\Bbb Z} w_2(M) = 0$.
A really elegant proof of this fact is given in the main proposition of this short note. I will not reproduce it. The essential point is that the Bockstein long exact sequence shows that $\beta_{\Bbb Z} w_2(M) = 0$ if and only if $w_2(M)$ lifts to an integral class, and that note explains how to show that $w_2(M)$ lifts to an integral class.