Cobordism and finite sheeted covers of manifolds

Question

Let $M$ be an oriented manifold, not necessarily compact. Let $M'$ be a (finite) $k$-sheeted cover and let $\pi:M'\longrightarrow M$ be the covering map.

Question 1 : Is it true that $M'$ is (oriented) cobordant to $k$ disjoint copies of $M$?

Question 2 : If the answer to the above is true then let $W$ be such a cobordism. Is there a map $\widetilde{\pi}:W\longrightarrow M$ such that $\widetilde{\pi}$ restricts to the natural maps on each end?

I would guess that the answers to these questions should be well-known. However, any references, proofs or counter-examples(?) would be helpful.

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EDIT : From the discussion/answers below, it is clear that the answer to (1) is positive when $M$ is a closed, oriented manifold. The characteristic numbers determine the oriented cobordism class of $M$. However, when $M$ is an open manifold, these numbers do not make immediate sense. I'm particularly interested in both questions in this scenario.

Answer 1

The answer is yes to the first question. This follows from the fact that characteristic numbers determine completely cobordism classes. Actually both questions have two versions: an oriented and an unoriented one.

The fact that the characteristic numbers are multiplied by $k$ under a $k$-fold covering follows from the definition of characteristic numbers through vector fields. (Each Pontrjagin class and Stiefel Whitney class can be obtained by taking some, properly chosen number of generic vector fields and consider the points, where the dimension of the space spaned by them drops by (at least) a given number compered with the possible maximum. (This given number is $1$ for $W$ and 2 for the $P$ classes.) The set of these points is a cycle dual to the corresponding characteristic class. Hence the characteristic numbers can be obtained as the number of intersections of such cycles. Now by the covering map we can pull back the vector fields, we can pull back the cycles, and clearly the number of intersections of the cycles in the covering manifold will be $k$ times that in the base manifold.

Answer 2

Here is Dold's argument showing that if $M' \to M$ is a double cover and an orientation preserving map, than $M'$ is oriented-cobordant to two copies of $M$.

Let $V \subset M$ be a codimension $1$ submanifold representing the $Z_2$-homology class dual to $W_1$ of the double covering $M' \to M.$ Let $N$ be the closed tubular neighbourhood of $V$ in $M.$ Note, that if we take two copies of $M \setminus int N$ and attach them appropriately along the boundaries, then we obtain $M'.$

Now take two copies of the cylinder $M\times [0,1]$ and attach them along $N \times 1$ to each other so that from the two copies of $M \times {1}$ we obtain $M'$. We obtain a compact, oriented manifold with boundary, the boundary consisists of two copies of $M \times {0}$ and one copy of $M'$

Answer 3

Its not true in complex cobordism. In Quillen's paper ``Elementary proofs of some results of cobordism theory using Steenrod operations", Section 4, he computes the complex cobordism class of a principal $\mathbb{Z}/k$-bundle $p : E \to B$ in terms of the associated line bundle $L \to B$ (given by $\mathbb{Z}/k \subset \mathbb{C}^\times$).

The answer is that $[p : E \to B] = [k](e(L))/e(L) \in MU^0(B)$, where $[k](x)$ is the $k$-series of the formal group law for complex cobordism. You can easily cook up examples, such as manifold approximations to skeleta of $B\mathbb{Z}/k$, for which this is not $k\cdot [\mathrm{id} : B \to B]$.

Presumably you wanted an answer in oriented cobordism. Perhaps the above counterexample survives under $MU^*(-) \to MSO^*(-)$?