# Transverse intersection is a well-defined operation on cobordism classes

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.