Computing the cohomology groups of the Klein bottle as a $\Delta$-complex I am currently working on how to compute the cohomology and ring structure of certain surfaces who are given as $\Delta$-complexes such as the Kein bottle pictured below.

For this i encountered this particular answer:
https://math.stackexchange.com/a/877083/543570
Unfortunately, i do not understand why it holds that
$\operatorname{Im}\delta:C^1 \rightarrow C^2=\langle \mu+\lambda, \mu- \lambda\rangle=\langle2\mu, \mu+\lambda\rangle=\langle\mu +\lambda\rangle$
and
$\ker \delta:C^1 \rightarrow C^2=\langle\beta+\gamma,\alpha +\beta\rangle$
where $\alpha,\beta,\gamma$ are the dual basis elements of $a,b,c$ respectively and $\mu,\lambda$ the dual basis elements for $U,L$ respectively.
Could someone help me understand why $\operatorname{Im}\delta = \langle\mu +\lambda\rangle$ and $\ker \delta = \langle\beta+\gamma,\alpha +\beta\rangle$ ?
Unfortunately i can't provide any own attempts since i got stuck at these two computations and do not know how to continue.
Thanks for any help!
 A: In the linked question, with $\alpha$ dual to $a$, etc., and $\mu, \lambda$ dual to $U, L$, the author says [I've added sequence numbers]


*

*To compute cohomology we need the homology groups of the chain complex: $0 \rightarrow C^0 \rightarrow C^1\rightarrow C^2 \rightarrow 0\rightarrow ...$, so we compute the coboundary map.





*$\delta(\phi)(a)=\phi(\partial(a))=\phi(v)-\phi(v)=0=\delta(\phi)(b)=\delta(\phi)(c)$, so $\delta:C^0 \rightarrow C^1$ is the zero map.





*$\delta(\alpha)(U)=\alpha(\partial(U))=\alpha(a)+\alpha(b)-\alpha(c)=1+0-0=1$.





*Similarly, $\delta(\beta)(U)=1$ and $\delta(\gamma)(U)=-1$. And $\delta(\alpha)(L)=1$, $\delta(\beta)(L)=-1$, $\delta(\gamma)(L)=1$.





*So $\delta:C^1 \rightarrow C^2$ takes $\alpha \rightarrow \mu +\lambda$, $\beta \rightarrow \mu -\lambda$, and $\gamma \rightarrow -\mu +\lambda$. Thus:





*Im $\delta:C^1 \rightarrow C^2=<\mu+\lambda, \mu- \lambda>=<2\mu, \mu+\lambda>=<\mu +\lambda>$ since we are over $\mathbb{Z}_2$.


Now let's recall that $a,b,c$ generate $C_1$, and $\alpha, \beta, \gamma$ generate $C^1$, etc. Item 4 says that $\delta(\alpha)$ can be determined, as an element of $C^2$, by what it does to generators (i.e.  $U$ and $L$) for $C_2$. From item 3, we have that $\delta(\alpha)(U) = 1$; item 4 says $\delta(\alpha(L)) = 1$. Now the question is "what linear combination of the basis elements $\lambda$ and $\mu$ also has these properties, for every element of $C^2$ is a combination of the basis elements. The answer is that $\lambda + \mu$ does. Hence
$$
\delta(\alpha) = \lambda + \mu.
$$
You can use the same process to figure out what delta does to the other two generators for $C^1$. When you're done, you've got three elements of $C^2$ which span the image of $\delta$. That's what items 5 says.
Item 6 then computes a reduced form of this span, using the fact that you're working over the integers mod 2 to get rid of some stuff.
So...exactly which of the items in the proof is confusing you?
Post-comment addition
$$\newcommand{\two}{{\Bbb Z/2\Bbb Z}}
$$
$C^1$ is a 3-dimensional vector space over $\two$, with basis $\alpha, \beta, \gamma$; $C^2$ is a 2-dimensional vectorspace over $\two$, with basis $\mu, \lambda$. The function $\delta$ is a linear transformation from the first to the second, and hence can be represented (in these two bases) by matrix multiplication with a $2 \times 3$ matrix, which is
$$
A = \pmatrix{
1 & 1& -1 \\
1 & -1& 1
}
$$
(where do each column's entries come from? Make sure you know!)
but since we're talking about mod-2 stuff, that might as well be
$$
A = \pmatrix{
1 & 1& 1 \\
1 & 1& 1
}
$$
The kernel of $T(v) = Av$ is the same as that of $T(v) = A'v$, where $A'$ is the row-reduced version of $A$, which is
$$
A' = \pmatrix{
1 & 1& 1 \\
0 & 0& 0
}
$$
which you get by subtracting the first row from the second. The first column (corresponding to $\alpha$) contains a leading "1" for some row, but the others do not; they can therefore be treated as free variables. So looking at $x\alpha + y\beta + z\gamma$, $y$ and $z$ are free variables. By assigning $y = 1, z = 0$, we find $x = 1$, i.e., that $\alpha + \beta$
is one generator for the kernel. By assigning $y = 0, z = 1$, we again find $x = 1$, so that $\alpha + \gamma$ is another generator for the kernel. So the kernel is spanned by these two elements. (Of course, there are many other pairs of elements that span it too).
A: Following the answer that you link, recall that the dual $\mu\in C^2(K,\mathbb{Z}_2)$ of $U$ is the linear map defined to be $1$ at $U$ and $0$ at $L$ (since $C_2(K,\mathbb{Z}_2)$ is spanned by $U$ and $L$). Similarly, $\lambda$ is the dual of $L$, so it is $1$ at $L$ and $0$ at $U$.
Image
In the answer, $\delta(\alpha)$, $\delta(\beta)$ and $\delta(\gamma)$ are computed, i.e. the image of the generators of $C^1$ are computed. In particular $\delta(\alpha)(U)=1=\delta(\alpha)(L)$. Since $\delta(\alpha)$ is a linear combination of $\lambda$ and $\mu$, and its values at $U$ and $L$ are both $1$, it follows that $\delta(\alpha)=\mu+\lambda$.
Kernel
An element of $C^1$ that belongs to the kernel of $\delta$ is a combination of $\alpha$, $\beta$ and $\gamma$ that is sent to $0$ by $\delta$. Since we have the images $\delta(\alpha)$, $\delta(\beta)$ and $\delta(\gamma)$, and since we are over $\mathbb{Z}_2$ we can simply check which combinations yield $0$. For instance, $\delta(\alpha)(U)=\delta(\beta)(U)$ and $\delta(\alpha)(L)=\delta(\beta)(L)$ (equalities in $\mathbb{Z}_2$), so $\delta(\alpha+\beta)=0$ because $\delta(\alpha+\beta)(V)=0$ for all $V\in C_2$, since $C_2$ is spanned by $U$ and $L$, both on which $\delta(\alpha+\beta)$ vanishes.
Can you follow from here?
