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!
Best Answer
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?