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

 A: $\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$
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.
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, 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.
