The precise statement on J. W. Morgan's "The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds (MN-44)" that 4-manifold $X$ admits a Spinc structure (Lemma 3.1.2) seems to be that [every 4-*orientable* manifold X admits a $Spin^c$ structure.](https://mathoverflow.net/questions/266946/every-4-manifold-has-a-spinc-structure) This means that we impose the $w\_1(X)=0$ for the 4-*orientable* manifold X.


However, for 4-*unorientable* manifold $M$, we may modify the statement to different structures.


In addition to,
$Spin^c=\frac{(Spin \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$


we can have:
$$Pin^c=\frac{(Pin^+ \times U(1))}{\mathbb{Z}/2\mathbb{Z}}=\frac{(Pin^- \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ 
$$Pin^{\tilde c+}=\frac{(Pin^+ \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ 
$$Pin^{\tilde c-}=\frac{(Pin^- \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}}.$$


See for example this [Ref, Annals of Physics 394, 244-293 (2018)](https://www.sciencedirect.com/science/article/pii/S000349161830112X?via%3Dihub) and References therein.


It seems that I can improve John Morgan's statement to show that 



> 
> Every unorientable 4-manifold has either a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure. (?)
> 
> 
> 


e.g. My approach is based on improving the map


\begin{equation\*}
H^1(X;Pin^c) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}\_2),
\end{equation\*}


\begin{equation\*}
H^1(X;Pin^{\tilde c+}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}\_2),
\end{equation\*}


\begin{equation\*}
H^1(X;Pin^{\tilde c-}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}\_2).
\end{equation\*}
The last maps of all three need to have appropriate constraints between $c\_1$ and $w\_2(M)$, $w\_1(M)$ and $w\_1^2(M)$.



> 
> *Question*: I wonder whether there exists any math literature show the similar results like mine above? Or are my statements obviously true? (Or obviously wrong?)
> 
> 
> 



Every unorientable 4-manifold has a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure
$\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$
The conjecture is false: $\RP^4\amalg(\RP^2\times\RP^2)$ is an unorientable 4-manifold that has no
pin$c$, pin$\tilde c-$, or pin$\tilde c+$ structure.


It suffices to find three unorientable 4-manifolds $A$, $B$, and $C$, such that $A$ isn't pin$c$, $B$
isn't pin$\tilde c-$, and $C$ isn't pin$\tilde c+$. Then $A\amalg B\amalg C$ doesn't admit
any of the three structures: a $G$-structure is a reduction of the principal bundle of frames, so a $G$-structure
on a manifold induces a $G$-structure on each connected component.


First, $A = \RP^2\times\RP^2$ doesn't have a pin$c$ structure; this is discussed
[here](https://mathoverflow.net/q/228166/97265).


Then, $B = \RP^4$ has no pin$\tilde c-$-structure. Let $H^\*(-; \Z\_{w\_1})$ denote integer cohomology
twisted by the orientation bundle. From [Shiozaki-Shapourian-Gomi-Ryu](https://arxiv.org/abs/1710.01886), Lemma
D.7, we learn that a manifold $M$ has a pin$\tilde c-$-structure iff $w\_2 + w\_1^2\in H^2(M;\Z/2)$ admits
a lift across the mod 2 reduction map


$$\tag{$\*$} H^2(M;\Z\_{w\_1})\longrightarrow H^2(M; (\Z/2)\_{w\_1}) = H^2(M;\Z/2).$$


Twisted Poincaré duality tells us that if $M$ is a closed $n$-manifold, there are isomorphisms $H\_k(M;\Z)\cong
H^{n-k}(M;\Z\_{w\_1})$, so $H^2(\RP^4;\Z\_{w\_1})\cong H\_2(\RP^4;\Z) = 0$. A quick computation shows that if $x\in
H^1(\RP^4;\Z/2)$ is the generator, then $w\_2(\RP^4) + w\_1(\RP^4)^2 = x^2$; in particular, it's nonzero, so it can't
lift to $H^2(\RP^4;\Z\_{w\_1})$. Therefore $\RP^4$ admits no pin$\tilde c-$-structure.


Finally, $C = \RP^2\times\RP^2$ doesn't have a pin$\tilde c+$-structure. Lemma D.8 of the
Shiozaki-Shapourian-Gomi-Ryu paper tells us that a manifold $M$ has a pin$\tilde c+$-structure iff
$w\_2\in H^2(M;\Z/2)$ admits a lift across ($\*$).


Let $x$ denote the generator of $H^1(-;\Z/2)$ of the first $\RP^2$ and $y$ be that for the second $\RP^2$. Using
the usual CW structure on $\RP^2$ and the product CW strucure on $\RP^2\times\RP^2$, you can check that
$H\_2(\RP^2\times\RP^2;\Z)\cong\Z/2$ and the reduction mod 2 map $H\_2(\RP^2\times\RP^2;\Z)\to
H\_2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H\_2(\RP^2\times\RP^2;\Z)$ to the Poincaré dual of $x+y$.
The Poincaré duality isomorphisms $H\_2(\RP^2\times\RP^2;\Z)\cong H^2(\RP^2\times\RP^2;\Z\_{w\_1})$ and
$H\_2(\RP^2\times\RP^2;\Z/2)\cong H^2(\RP^2\times\RP^2;\Z/2)$ are natural with respect to change-of-coefficients,
which means that the reduction mod 2 map ($\*$) for $M = \RP^2$ sends the nonzero element of
$H^2(\RP^2\times\RP^2;\Z\_{w\_1})\cong\Z/2$ to $x+y$. However, $w\_2(\RP^2\times\RP^2) = x^2+xy+y^2$, so it's not in
the image of ($\*$), and therefore $\RP^2\times\RP^2$ has no pin$\tilde c+$-structure.


