# Spanier-Whitehead dual and Hopf fibration

Consider a map of spheres $f:S^n\to S^m$ covered by a map of trivial $\mathbb R^k$-bundles. In other words, we take the trivial rank $k$ vector bundle over $S^m$ and pull it to $S^n$ via $f$. Consider the corresponding map of the Thom spaces, and its Spanier-Whitehead dual. How is the dual related to $f$?

The test case I care about is when $f$ is the Hopf fibration $S^3\to S^2$. Then I think the dual can be represented by a map $F:\Sigma^{r+1}(S^2_+)\to\Sigma^r (S^3_+)$ where $X_+$ means $X$ disjoint union a point, and $\Sigma^s$ is $s$-fold suspension. Thus $F$ can be thought of as a map $S^{r+3}{\vee} S^{r+1}\to S^{r+3}{\vee} S^r$. What is this map?

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(I'm going to assume you only care about the homotopy class of map.)

Such maps are roughly their own duals. The Spanier-Whitehead dual of a map $S^n \to S^m$ is a map $"S^{-m}\_+ \to S^{-n}\_+"$, which is represented by $\Sigma^{r-m} S^m\_+ \to \Sigma^{r-n} S^n\_+$ for sufficiently large r. If r=n+m+s, this is a map $\Sigma^{s+n} S^m\_+ \to \Sigma^{s+m} S^n\_+$. This is homotopy equivalent to a map $S^{n+m+s} \vee S^{n+s} \to S^{n+m+s} \vee S^{m+s}$.

The resulting map is the wedge of the identity map on the first factor and the (up to sign, which I will get wrong if I try) the s-fold suspension of $f$ on the second factor.

This is easier to say in the stable homotopy category, where your original map become stably the map $id \vee \Sigma^\infty f:S^0 \vee S^n \to S^0 \vee S^m$ and the dual just dualizes the maps on each factor. It's also easier to say if you use based maps of spheres in the first place.

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I am puzzled that you are getting a self-map of a wedge of spheres while in my question it is a map of wedges of spheres of different dimensions. In other words, stably the dual is a map from the suspension of $S^m_+$ to $S^n_+$, not from $S^m_+$ to $S^n_+$. – Igor Belegradek Nov 19 '09 at 20:36
I'm confused, I don't see any maps which are self-maps - the closest is the map $\Sigma^{s+n} S^m_+ \to \Sigma^{s+m} S^n_+$, which have the same dimension but the basepoints suspend to cells of different dimension. – Tyler Lawson Nov 19 '09 at 20:55
Sorry I misspoke about self-maps. What I do not understand is that you seems to be saying that $\Sigma^{n+s}(S^m\vee S^0)\to \Sigma^{m+s}(S^n\vee S^0)$ stably equals to $S^m\vee S^0\to S^n\vee S^0$. Is this correct? – Igor Belegradek Nov 19 '09 at 21:14
Also you seem to be using that "dual of the wedge is wedge of the duals". Where can I find this stated or proved? Thanks! – Igor Belegradek Nov 19 '09 at 21:23
That map should instead stably be equal to the map $S^{-m} \wedge S^0 \to S^{-n} \wedge S^0$. I think that this material should be covered in J. F. Adams' Chicago notes on stable homotopy theory - roughly the wedge is both product and coproduct in the stable homotopy category, and so is preserved by this dualization. – Tyler Lawson Nov 19 '09 at 22:36