TITLE: Every 4-manifold has a $Spin^c$ Structure
QUESTION [10 upvotes]: 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:

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 $X$?
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.

REPLY [3 votes]: 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.
Now to your actual questions:

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.)

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.