Measuring the failure of pushforward to commute with Steenrod squares

Question

Let $f \colon X \rightarrow Y$ be a map of topological spaces. Let's say that they are (closed) manifolds (not necessarily orientable), though to be honest I'm really interested in the more general setup of $\mathbb{Z}/2$-Poincare duality spaces.

(From now on all cohomology coefficients are $\mathbb{Z}/2$.) Poincare duality induces, from the usual pullback, a pushforward map

$$ f_* \colon H^*(X) \rightarrow H^*(Y) .$$

While pullback commutes with Steenrod square, the pushforward unfortunately does not.

Question: can one "quantify" the failure of this commutativity? (i.e. some nice expression for the difference $f_* Sq - Sq f_*$?)

I have an answer to my question in terms of Wu classes, but I want something that doesn't involve them. Here's an example of the kind of thing that I'm looking for.

Suppose now that $X \hookrightarrow Y$ is a closed embedding of smooth manifolds. A reliable friend told me the following nice formula measuring the failure of this commutativity. If $N_{X/Y}$ is the normal bundle of $X$ in $Y$, then by the Gysin map, tubular neighborhood, excision etc. we get a composition

$$H^*(X) \xrightarrow{\text{Gysin}} H^{*+r}(N_{X/Y}, N_{X/Y}-X) \cong H^*(Y, Y-X) \xrightarrow{g^*} H^*(Y)$$

and the claim is that this composition is $f_*$. In particular, this can be used to write down a formula for the "error term" in commuting Sq with pushforward: if $u$ is the Thom class of $N_{Y/X}$ then

$$ f_*(Sq \alpha) = g^* (Sq \alpha \cup u) = Sq (g^* \alpha \cup Sq^{-1} u) $$

Now by definition $Sq^{-1} u = w^{-1} u$ where $w$ is the (total) Stiefel-Whitney class of $N_{Y/X}$, so

$$ f_* (Sq \alpha) = Sq ( g^* \alpha \cup w^{-1} u ) = Sq (f_*(\alpha \cup w^{-1})) $$

Unfortunately this doesn't make literally make sense if $X,Y$ are not smooth (since there's no $N_{X/Y}$). However, some things make sense; for instance, we can say that $g^* u = f_*(1)$. In my situation $f_*(1)$ is very simple; for instance I know that $Sq (f_*(1)) = f_*(1)$. (This is basically why I suspect that ``$w"=1$ in my case, so that $f_*$ and Sq should commute.)

Question: is there a "salvage" of the above formula to the case where $X,Y$ are not smooth?

I would prefer an answer that is formulated at the level of Poincare duality spaces, since that's more like my situation (and the question makes sense at that level).

Answer 1

If $f: X \longrightarrow Y$ is a map whose homotopy fibers are all Poincaré duality spaces then one gets a wrong-way map in the category of spectra: $\Sigma_+^{\infty}Y \longrightarrow X^{\nu}$ where $\nu$ is the stable spherical fibration given by the Spivak normal bundle of this map. On cohomology we get $g: H\mathbb{F}_2^*(X^{\nu}) \longrightarrow H\mathbb{F}_2^*(Y)$ which commutes with all Steenrod operations since it comes from a map of spectra. Composing with the Thom isomorphism $H\mathbb{F}_2^*(X) \cong H\mathbb{F}_2^*(X^{\nu})$ (where there's a degree shift) gives your umkehr map coming from Poincaré duality. The Thom isomorphism comes from cupping with the Thom class $U_{\nu}$, so you could define some notion of Stiefel-Whitney class by $Sq^{-1}U_{\nu}$ and get a formula similar to the one you've got for an embedding of smooth manifolds.

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Here's a sketch of a construction of that pushforward map. If $Y$ is a point we'd like a stable spherical fibration $\nu$ on $X$ together with a map $S^0 \longrightarrow X^{\nu}$ of spectra sending the generator in homology to the fundamental class under the Thom isomorphism (which we can arrange if $X$ is simply connected so that $\nu$ is orientable). Such a thing exists and is unique up to fiberwise stable homotopy and it's called the Spivak normal fibration. You can also get something when $X$ isn't simply connected, and you can arrange for these to vary in a family over a base $Y$ and get the map I've indicated.

In the case of a map of manifolds this goes back at least to Atiyah: factor $f$ as $X \hookrightarrow \mathbb{R}^N \times Y \longrightarrow Y$ and let $\nu_N$ be the normal bundle to the first inclusion. Then Pontryagin-Thom collapse gives a map $S^N \wedge Y_+ \longrightarrow X^{\nu_N}$, and then let $N \rightarrow \infty$ to get the map of spectra above.