1
$\begingroup$

I just wanted to double check that I'm not making a mistake here. For the $0$th cohomology group $H^{0}(X;G)$ of a space $X$, we can think of the elements as being functions $X \rightarrow G$ that are constant on path components of $X$.

Hatcher says that the reduced cohomology $\tilde{H^0}(X;G)$ is all functions $X \rightarrow G$ that are constant on path components, modulo the functions that are constant on all of $X$.

I was thinking that we might as well let $g: X \rightarrow G$ be the function that sends all of $X$ to $g \in G$, and then from his description $\tilde{H}^0(X;G) \simeq H^0(X;G)/G$, is this correct?

$\endgroup$

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.