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?

  • 1
    $\begingroup$ 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 $\partial\circ\partial = (\partial_\alpha \circ \partial_\alpha) = 0$. $\endgroup$ – jgon May 16 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.