possibly simple exercise in algebraic topology regarding Bockstein homomorphism Sorry for the uninformative title; I just can't choose a good title that explains my question.
I'm reading Manolescu's lectures on the triangulation conjecture. What I'm stuck is exercise 2.4, which asks to show:
For a manifold $M$ whose dimension is $\geq 5$, if $Sq^1(\Delta(M))\neq 0$, then $\delta(\Delta(M))\neq 0$. 
Now I should explain the notations here before prospective problem solvers get lost and go away; Here, $\Delta(M)\in H^4(M;\mathbb{Z}/2\mathbb{Z})$ is something called the Kirby-Siebenmann class. Its definition may not matter; I'll give some properties of this obstruction class in a moment. And $Sq^1$ is obviously the first Steenrod operator, which is the same as the Bockstein homomorphism $\beta:H^\bullet(M;\mathbb{Z}/2\mathbb{Z}) \to H^{\bullet+1}(M;\mathbb{Z}/2\mathbb{Z})$. And $\delta$ is the connecting homomorphism of the following LES:
$\cdots H^4(M;\Theta^\mathbb{Z}_3)\overset{\mu}{\to} H^4(M;\mathbb{Z}/2\mathbb{Z})\overset{\delta}{\to} H^5(M;\mathrm{ker} \mu)\to \cdots$
which is associated to the SES $0\to \mathrm{ker}\mu \to \Theta^\mathbb{Z}_3\overset{\mu}{\to} \mathbb{Z}/2\mathbb{Z}\to 0 $.
Do not care about $\mu$ or $\Theta^\mathbb{Z}_3$ if you haven't heard; I guess the only important part is that the SES above does not split. 
Now I give some properties that is (supposed to be) relevant to this question.
$\delta(\Delta(M))=0$ implies $\Delta(M)$ is the image of an element in $ H^4(M;\Theta^\mathbb{Z}_3)$ under $\mu$, ie, $\Delta(M)=\mu(c)$ for some $c\in H^4(M;\Theta^\mathbb{Z}_3)$. 
(As stated above,) the SES $0\to \mathrm{ker}\mu \to \Theta^\mathbb{Z}_3\overset{\mu}{\to} \mathbb{Z}/2\mathbb{Z}\to 0 $ does not split.
(might not be relevant) $\Theta^\mathbb{Z}_3$ is infinitely generated and has a $\mathbb{Z}$-summand.

Any help would be appreciated; I'm sure this should be a simple algebraic topology exercise, but somehow got stuck after spending few hours on it.
 A: Note that for any abelian group $A$, there is a natural isomorphism $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z},A)\cong A/2A$.  Indeed, this is just what you get by computing this Ext using the resolution $0\to\mathbb{Z}\stackrel{2}\to\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0$ of $\mathbb{Z}/2\mathbb{Z}$.  In particular, let's take $A=\ker\mu$ and consider the class $x\in\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z},A)\cong A/2A$ representing the extension $$0\to A \to \Theta^\mathbb{Z}_3\overset{\mu}{\to} \mathbb{Z}/2\mathbb{Z}\to 0.$$  Since $A/2A$ is a vector space over $\mathbb{Z}/2\mathbb{Z}$ and $x$ is nonzero (the SES does not split), there is a homomorphism $A/2A\to\mathbb{Z}/2\mathbb{Z}$ which sends $x$ to $1$.  Composing this with a quotient map $A\to A/2A$, we get a homomorphism $f:A\to\mathbb{Z}/2\mathbb{Z}$.  This homomorphism has the property that the induced map $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z},A)\to\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})$ sends $x$ to the nonzero element of $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})$, which represents the extension $$0\to\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0.$$  This implies that there is a homomorphism $g$ filling in the following diagram:
$$\require{AMScd}
\begin{CD}
0 @>>>A@>{}>> \Theta^\mathbb{Z}_3 @>{\mu}>> \mathbb{Z}/2\mathbb{Z} @>>>0\\
& @V{f}VV @V{g}VV @V{1}VV \\
0 @>>>\mathbb{Z}/2\mathbb{Z}@>{}>> \mathbb{Z}/4\mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z} @>>>0\\
\end{CD}$$
(Explicitly, if we choose an element $y\in A$ whose image in $A/2A$ is $x$, then we can identify $\Theta^\mathbb{Z}_3$ with the set $A\times\mathbb{Z}/2\mathbb{Z}$ with the group operation which is coordinatewise except that $(0,1)+(0,1)=(y,0)$.  The map $g$ then sends $(a,i)\in A\times\mathbb{Z}/2\mathbb{Z}$ to $2f(a)+i$.  This works because we chose $f$ so that $f(y)=1$.)
This map of short exact sequences then induces maps between the associated long exact sequences in cohomology.  In particular, the induced maps on cohomology commute with the Bocksteins, so the induced map $f_*:H^5(M;A)\to H^5(M;\mathbb{Z}/2\mathbb{Z})$ satisfies $f_*(\delta(c))=Sq^1(c)$.  Taking $c=\Delta(M)$ gives your desired result.
