Prove that $\{C_n\}_{n\in\mathbb{Z}}$ is a chain complex with homomorphism border $\partial:=\bigoplus_{\alpha\in\Lambda}\partial^{\alpha}$.

Let $$\Lambda$$ be a fixed set. For each $$\alpha\in\Lambda$$ is $$\{C_n^{\alpha}\}_{n\in\mathbb{Z}}$$ a complex of chains with homomorphism border $$\partial^{\alpha}$$.

For each $$n\in\mathbb{Z}$$ we define $$C_n=\bigoplus_{\alpha\in\Lambda}C_n^{\alpha}$$. Prove that $$\{C_n\}_{n\in\mathbb{Z}}$$ is a chain complex with homomorphism border $$\partial:=\bigoplus_{\alpha\in\Lambda}\partial^{\alpha}$$.

To show that $$\{C_n\}_{n\in\mathbb{Z}}$$ is a complex of chains with homomorphisms border $$\partial$$. We have to prove that $$\partial\circ\partial=0$$. For this, let's take $$(x_{\alpha})\in C_n$$ so $$(\partial\circ\partial)(x_{\alpha})=\partial(\partial(x_{\alpha}))=\partial((\partial^{\alpha}(x_{\alpha})))=(\partial^{\alpha}(\partial^{\alpha}(x_{\alpha})))=((\partial^{\alpha}\circ\partial^{\alpha})(x_{\alpha}))=(0)$$. Is this demonstration okay? What can be improved? Is there another way to do this?

• That's fine. If you know that direct sum is a functor, then there's no need to introduce $(x_\alpha)$. You can just say $\partial\circ\partial = (\partial_\alpha \circ \partial_\alpha) = 0$. – jgon May 16 at 18:26