I am reading the book by Milnor & Stasheff on Characteristics Classes, In the 4th Chapter There's is this beautiful result by Thom" If all of the Stiefel-Whitney numbers of M are zero, then M can be realized as the boundary of some smooth, compact manifold."
Just after which as an example It is written that "A disjoint union of two copies of a manifold M has all Stiefel-Whitney no. zero."
Why is that true? It can easily be seen by applying the converse of above as the disjoint union can be regarded as the boundary of M x [0,1]. Is there a way to directly prove it?
Any hints are appreciated.
It is clear from the proof of additivity of cohomology under disjoint union that $H^*(M \sqcup M) = H^*(M) \oplus H^*(M)$ as rings.
It is also clear from the definition of the Stiefel-Whitney classes (for instance, naturality) that $w_i(M \sqcup M) = (w_i(M), w_i(M))$ above.
For a partition $p$ of $n = \dim M$ (a sequence of integers $1 \leq i_1 \leq \cdots \leq i_j \leq n$ with $i_1 + \cdots + i_j = n$) there is an associated product $$w_p(M) = w_{i_1}(M) \cdots w_{i_j}(M) \in H^n(M).$$
In your case, the first two sentences give you that $w_p(M \sqcup M) = (w_p(M), w_p(M)) \in H^n(M \sqcup M) = H^n(M) \oplus H^n(M)$.
Now the Stiefel-Whitney numbers of $M$ are obtained by pairing with the fundamental class; I write $s_p(M) = \langle w_p(M), [M]\rangle$ for this integer mod 2.
Then we have $$s_p(M \sqcup M) = \langle w_p(M \sqcup M), [M \sqcup M]\rangle = \langle (w_p(M), w_p(M)), ([M], [M])\rangle = \langle w_p(M), [M]\rangle + \langle w_p(M), [M]\rangle = 2s_p(M) = 0.$$