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I found a few occurrences of the same proof about representability of elements of $H_2(M,\mathbb{Z})$ for $M$ a closed orientable smooth $4$-manifold. All of them stop at the very end claiming that the conclusion is clear. I found an analogue question on MO but even there it's not explained the last passage of the proof.

(from Kirby's Topology of $4$-Manifolds Theorem 1.1 page $20$) There is an isomorphism $$ H^2(M, \mathbb{Z})\cong [M,\mathbb{C}P^{\infty}]$$ so letting $\hat{\alpha}$ being the Poincaré dual of a chosen $\alpha \in H_2(M, \mathbb{Z})$, there is an homotopy class of maps $[f]\colon M \to \mathbb{C}P^{\infty}$ corresponding to $\hat{\alpha}$. By cellular approximation, we can homotopy (a representative of) $f$ in roder to obtain a map $f\colon M \to \mathbb{C}P^2$. Make $f$ smoothly transverse to $\mathbb{C}P^1\subset \mathbb{C}P^2$. Consider $f^{-1}(\mathbb{C}P^1)$, this will be an oriented surface representing $\alpha$.

I'm not able to prove this last statement, everything is pretty much clear. Looking around I didn't find any explanation for this, so I'm wondering if it is a trivial result. The only indication I found is in the linked question where one suggests to sue Pontrjagin-Thom construction for the group $SO(2)$. Needless to say I'm unable to make use of this hint.

Can someone give me an explanation for this last sentence?

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Luigi M
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1 Answers1

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The reason for this is the following fact:

Theorem: Let $N\to M$ be a submanifold and $[N]$ the homology class it defines in $M$. Then in $M$ the Thom class of the normal bundle is dual to $[N]$. (see e.g. this exercise in Milnor Stasheff)

Use this together with the map $[M,K(G,i)] \to H^i(M;G)$ which is precisely given by pulling back the unique $i$-cohomology class $k$.

Third thing we use is that normal bundles pull back and naturality of characteristic classes.

Now in your notation (let additionally $\tau_\alpha \in H^2M$ and $\tau_k\in H^2\mathbb CP^2$ be the Thom classes of the normal bundles of $f^{-1}(\mathbb CP^1) $ and $\mathbb CP^1$ resp.):

$$ \hat \alpha =f^*k = f^*(PD(\mathbb CP1)) = f^*(\tau _k) = \tau_\alpha. $$

Apply PD to this equality and obtain $$\alpha = PD(\hat \alpha)=PD(\tau_\alpha)=[f^{-1}(\mathbb CP^1)]$$.

Sorry for the very confusing notation which already kind of assumes the result.

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Daniel Valenzuela
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  • Thanks for the answer! could you explain why $k=PD([\mathbb{C}P^1])$? I mean, as far as I know, the bijection is given by pulling back the fundamental class defined as the preimage via UCT of the Hurewicz isomorphism. Is it trivial to see that this fundamental class correspond to the dual of the orientation class of $\mathbb{C}P^1$? – Luigi M Apr 13 '16 at 15:20
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    I mean in our case that doesn't really matter so much, as up to sign there's not really a choice ($Aut(\mathbb Z) = {\pm 1}$). If you worry about the sign (which corresponds to the change of orientation of the submanifold, and hence would still prove your original question about representability of homology classes), we would actually need to dig out your sign conventions and compute this. – Daniel Valenzuela Apr 13 '16 at 15:42