Every 4-manifold has a $Spin^c$ Structure

I'm having trouble understanding the proof given in Morgan's The Seiberg-Witten Equations that every 4-manifold $X$ admits a $Spin^c$ structure (Lemma 3.1.2). One can easily see from the exact sequence: \begin{equation*} H^1(X;Spin^c) \to H^1(X; SO(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{c_1+w_2} H^2(X;\mathbb{Z}_2) \end{equation*} that a $Spin^c$ structure will exist iff $w_2(TX)$ lifts to an integral class, which we can check using Bockstein homomorphisms. After that, I'm lost; I'm not sure if these are theorems, or whether they are supposed to be obvious:

1. In what sense is every $\mathbb{Z}/2^k \mathbb{Z}$ 3-class represented by a mapping from a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$?
2. Why are integral 2-classes that represent torsion elements necessarily represented by embedded oriented surfaces?

EDIT: Since the proof from Morgan's book is quite short, I may as well write out the whole thing here:

"We need only see that $w_2(X)$ lifts to an integral class $c \in H^2(X;\mathbb{Z})$ in order to prove the existence of a $Spin^c$ lifting. But for any class $x \in H_2(X;\mathbb{Z}/2 \mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two. To see that $w_2(X)$ lifts to an integral class, we must see that its integral Bockstein $\delta w_2(X)$ is zero. But this torsion integral class is zero if and only if it evaluates trivially on every $\mathbb{Z}/2^k \mathbb{Z}$ class of dimension three. Any such class is represented by a mapping of a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$. The value of $\delta w_2(X)$ on such a class is equal to the value of $w_2(X)$ on the Bockstein of this $\mathbb{Z}/2^k\mathbb{Z}$-manifold. Thus, we need only see that $w_2(X)$ vanishes on integral classes which represent torsion elements in $H_2(X;\mathbb{Z})$. But this is clear, any such class is represented by a smoothly embedded oriented surface with zero self-intersection".

I suppose what I'm really asking is which statements in this proof are non-trivial theorems about the topology of 4-manifolds, and which ones are supposed to be obvious?

Sorry, I don't seem to be able to comment, so I'll just say here: Ryan Budney, I hope this makes the question less vague, and Anton Fetisov, yes, there are other proofs of this fact that I do understand, but I'm specifically trying to understand this proof, because it seems very slick.

• Dimension 4 appears because you end up using Poincare duality (cup/cap product for computing intersection numbers). It is a classical result of Thom (google "representable classes") that every homology class in sufficiently low degrees (including 2 and 3) is represented by a submanifold, i.e. the fundamental class of such a submanifold is your desired homology class (explaining Question 1), and orientability for Question 2 is clarified for example in Torsten's answer here[mathoverflow.net/questions/22473/…. – Chris Gerig Apr 11 '17 at 23:31
• Thanks for the comment. Yes, I am familiar with the result of Thom, but I'm not aware of its extension to manifolds with a G-structure in Q1. Torsten's answer is nice but it seems to indicate that we didn't need the torsion condition at all in Q2. – jdk3264 Apr 12 '17 at 1:32
• Your question is rather vague. Could you write it up in a way that doesn't require readers to go somewhere else? – Ryan Budney Apr 12 '17 at 5:06
• Dimension 4 is required because any proof rests either on the Poincare duality (in compact case) or on the intersection form on 2-cohomology in general. Note that you also need orientability. I didn't read the paper you cite, but perhaps this paper will help. – Anton Fetisov Apr 12 '17 at 21:07
• Another proof along these lines is given in a short note of Teichner and Vogt, people.mpim-bonn.mpg.de/teichner/Papers/spin.pdf. You might find the exposition there helpful. – Danny Ruberman Apr 14 '17 at 14:12

First let me stress the importance of Anton Fetisov's comment that $X$ must be orientable. Indeed, this seems to be used at nearly every step of the proof you cite.
For example, consider the claim "for any class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two." This can fail if $X$ is non-orientable. For example, take $X=\mathbb{R}P^2\times\mathbb{R}P^2$, which has $w_2(X)=a^2\times 1 + a\times a + 1\times a^2$ where $a\in H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$ is the generator. Consider the homology class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ represented by the embedding $\iota:\mathbb{R}P^2\hookrightarrow \mathbb{R}P^2\times\mathbb{R}P^2$ of the first factor. Then $$\langle w_2(X),x\rangle =\langle w_2(X),\iota_*[\mathbb{R}P^2]\rangle = \langle \iota^*w_2(X),[\mathbb{R}P^2]\rangle = \langle a^2,[\mathbb{R}P^2]\rangle \equiv 1,$$ whereas it is clear that the self-intersection is zero in this case.
1. Here a $\mathbb{Z}/2^k\mathbb{Z}$-manifold means a manifold which is oriented with respect to singular homology with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients. When $k=1$ this is all manifolds, and the claim reduces to Thom's positive answer to the mod $2$ Steenrod realizability problem. When $k>1$, a $\mathbb{Z}/2^k\mathbb{Z}$-manifold is an orientable manifold $M^m$ with a choice of fundamental class $[M]\in H_m(M;\mathbb{Z}/2^k\mathbb{Z})$. By Thom's results, a homology class $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold if and only if its Poincaré dual in $H^1(X;\mathbb{Z}/2^k\mathbb{Z})$ is induced from the universal Thom class $t_{\mathbb{Z}/2^k\mathbb{Z}}\in H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})$ under some map $f: X\to MSO(1)$. Now note that $MSO(1)\simeq S^1$, and that the Thom class induces an isomorphism $$t^*_{\mathbb{Z}/2^k\mathbb{Z}}:H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})\cong H^1(K(\mathbb{Z}/2^k\mathbb{Z},1);\mathbb{Z}/2^k\mathbb{Z}).$$ It follows that every $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold.
(Note that orientability of $X$ was used here, to get Poincaré duality with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients.)
1. Again by orientability and Thom's results, every class in $H_2(X;\mathbb{Z})\cong H^2(X;\mathbb{Z})$ is represented by an embedded oriented surface $\Sigma\hookrightarrow X$. The torsion assumption is just used to get the final clause, that $\Sigma$ has zero self-intersection. This is because the self-intersection is Poincaré dual to cup square, and the square of a torsion class in $H^2(X;\mathbb{Z})$ is a torsion class in $H^4(X;\mathbb{Z})\cong \mathbb{Z}$, therefore is zero.