Let $M$ be a smooth manifold of dimension $n$, $\textsf{FC}_p (M)$ the set of all codimension $p$ framed cobordism classes of $M$. Just as a technical note, we are only considering compact submanifolds without boundary. I want to show that transverse intersection \begin{equation*} \pitchfork: \textsf{FC}_p (M) \times \textsf{FC}_q (M) \to \textsf{FC}_{p + q} (M) \end{equation*} is well-defined, i.e. for every $X \sim U$ codimension $p$ and $Y \sim V$ codimension $q$ framed submanifolds of $M$ satisfying $X \pitchfork Y$ and $U \pitchfork V$, we have $X \cap Y \sim U \cap V$.

My work so far: The natural first thing to try is let $Z, W \subseteq M \times [0, 1]$ be compact so that $\partial Z = X \sqcup U$ and $\partial W = Y \sqcup V$, though even if we perturb our $Z$ and $W$ so that $Z \pitchfork W$, it is not necessarily true that $\partial (Z \cap W) = (X \cap Y) \sqcup (U \cap V)$. Consider for example two "bended" cylinders (e.g. two coffee cup handles) so that they do not intersect in the middle.

Another approach that I was thinking of was using the correspondence $\textsf{FC}_p (M) \cong [M, S^p]$ set of smooth maps $f: M \to S^p$ quotiented by homotopy. We know that every framed manifold is a Pontryagin manifold, e.g. $X = f^{-1} (a)$ for a regular value $a$. My thought was that we look at what transverse intersection looks like on the smooth maps. I would think it looks something like \begin{equation*} \pitchfork: (f, g) \mapsto f \times g. \end{equation*} Then if $X = f_1^{-1} (a)$, $Y = g_1^{-1} (b)$, $U = f_2^{-1} (c)$ and $V = g_2^{-1} (d)$, clearly we have a homotopy from $f_1 \times g_1$ to $f_2 \times g_2$ since $f_1 \sim f_2$ and $g_1 \sim g_2$, and moreover $X \cap Y = (f_1 \times g_1)^{-1} (a, b)$ and $U \cap V = (f_2 \times g_2)^{-1} (c, d)$.

The issue is of course $f \times g$ is a map into $S^p \times S^q$, not $S^{p + q}$, but this technicality can be looked over if we can find an injective smooth map $\phi: S^{p} \times S^q \to S^{p + q}$, in which case we just post compose the product maps with $\phi$.


The map $\phi$ you are looking for is not injective. It is the map which crushes $S^p \vee S^q$ to a point. The assumption that $f$ is transverse to the basepoint of $S^p$ and similarly for $g$ implies that $(f,g)$ intersects $S^p \vee S^q$ transversely.


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