I have a question about an argument that occurred in the discussion about consequences of Bott periodicity in A Concise Course in Algebraic Topology by P. May on page 221. Here is the excerpt:
Firstly: we denote by $TO(q)$ the Thom space of the universal $q$-bundle $\gamma_q: E_q \to G_q(\mathbb{R})=BO(q)$.
In the excerpt May constructs a map $Tf_M \circ t: S^{n+q} \to TO(q)$ where $f_M: M \to BO(q)$ is the classifying map.
My question is why is this map cobordant? Namely why does it respect the cobordism relations?
Especially that would mean that if we have two $n$-manifolds $M,N$ and a $n+1$-manifold $W$ with boundary $\partial W =M \coprod N $ then the map $Tf_W \circ t$ should be nullhomotopic.
Intuitively I think the author means that the intuition behind it is that the homotopy is reached by "going along to inner" of $W$ from $M$ to $N$ but honestly it's to sloppy for me. I'm not sure why it should work globally.
Could anybody explain the argument and why it works ?
This is the Pontryagin-Thom construction, which is a general mechanism for showing that a cobordism theory is given by homotopy groups of a Thom spectrum. I'll talk about some of the aspects but leave a lot of details out.
It can be easier to treat $\mathcal{N}_n$ as the embedded cobordism groups, where each closed manifold $M^n$ is assumed to be embedded in $\mathbb{R}^\infty$, and a cobordism between $M$ and $N$ is a $W^{n+1}$ embedded in $\mathbb{R}^\infty \times [0,1]$ such that $W|_0 \cong M$ and $W|_1\cong N$. You can show using the Whitney embedding theorem this is the same as considering abstract manifolds without embeddings.
Given an $M\subset \mathbb{R}^\infty$, by compactness $M$ is actually contained in $\mathbb{R}^{n+k}$ for some finite $k$, with a $k$-dimensional normal bundle $\nu$. If we pick $k$ high enough then the embedding is unique up to isotopy and so $\nu$ is unique up to isomorphism. The bundle $\nu$ admits a bundle map $\varphi\colon\nu \to \gamma_k$ covering some $f\colon M \to BO(k)$, and so there is an induced map on Thom spaces $Tf\colon T\nu \to T\gamma_k$. If we choose an open tubular neighbourhood $U$ of $M$ in $\mathbb{R}^{n+k}$, or equivalently in $S^{n+k}$, we get a collapse map $$c\colon S^{n+k} \to S^{n+k}/U^c \cong D(\nu)/S(\nu)=T\nu$$ In other words, for a chosen embedding $M\subset \mathbb{R}^{n+k}$ we get a continuous function $S^{n+k}\to T\gamma_k$ given by $Tf\circ c$, and if we pick $k$ high enough its homotopy class doesn't depend on the embedding. If we picked another $k'$ we would get a continuous function between different spaces $S^{n+k'}\to T\gamma_{k'}$ but the idea is we end up getting an element of the limit $$[Tf\circ c]\in\pi_n MO = \lim_{k\to\infty} \pi_{n+k}T\gamma_k$$
Given a cobordism $W$ from $M$ to $N$, again we can pick some $k$ so that $W$ actually lives in $\mathbb{R}^k\times[0,1]$, and collapsing a tubular neighbourhood gives a map $Tf_W\circ c_W\colon S^{n+k}\times [0,1] \to T\gamma_k$ which is a homotopy between $Tf_M\circ c_M$ and $Tf_N\circ c_N$. This is roughly the sense in a cobordism of manifolds produces a homotopy of functions.
On the other hand, if $H\colon S^{n+k}\times [0,1] \to T\gamma_k$ is a homotopy, we can assuming wlog that it is transverse to the $0$-section of $\gamma_k$, i.e. the smooth manifold $BO(k)$, and then take the transverse intersection to produce a manifold $W$ of dimension $n+1$. The boundary is divided into two closed $n$-manifolds, one is $M=H_0^{-1}(BO(k))$ and the other is $N=H_1^{-1}(BO(k))$. This is roughly the way in which a homotopy of maps $S^{n+k}\to T\gamma_k$ produces a cobordism.