# Hopf invariant

1. Jun 9, 2010

### eok20

Can someone give me some hints on how to prove the following statement: if $$f: S^3 \to S^2, g: S^3 \to S^3$$ then $$H(f\circ g) = deg~g H(f)$$ where H is the Hopf invariant and deg g is the degree of g. I'm pretty clueless on how to start and I don't see how to get the deg to come in since that has to do with 3-d cohomology but the Hopf invariant has to do with 2-d and 4-d cohomology.

Thanks.

2. Jun 9, 2010

3. Jun 9, 2010

### eok20

Thanks-- I should have clarified. I'm interested in the algebraic topological definition and proof. Here, the Hopf invariant is the integer h such that $$\alpha \smile \alpha = h\beta$$ where $$\alpha$$ generates the 2-d cohomology and $$\beta$$ generates the 4-d cohomology of the mapping cone of $$f:S^3 \to S^2$$. The degree of $$g:S^3\to S^3$$ is the integer d such that $$f_* \gamma = d\gamma$$ where $$\gamma$$ generates the 3-d homology of S^3.

Unfortunately, I have very little intuition/feel for algebraic topology and I much prefer the differential forms and differential topology analogs.