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I am currently preparing for my exam of differential geometry and tried to solve the following question: Suppose $f,g : S^3 \to S^2$ are smooth maps. Do their induced maps (at the level of de Rham cohomology) agree?

I know that if I could show that the maps $f,g$ are smoothly homotopic that this would imply that the induced maps agree. However I have no idea how to do this (if it is even possible).

Thank you in advance.

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Have you tried writing down the de Rham cohomology groups of both spaces? How many different maps are possible between these groups? Can they all be induced by smooth maps?

Hint: $H^0_{dR}$ is the only tricky one.

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    So the de Rham cohomology groups in the case of $S^m$ are all zero except for $H^0$ and $H^m$, which are both isomorphic to $\mathbb{R}$. Hence the only possibilities for $H^1(S^2) \to H^1(S^3)$ are the zero map, for $ \mathbb{R} \cong H^2(S^2) \to H^2(S^3) = \{0\}$ also the zero map. But I don't really see what to do with the zeroth cohomology group, since both are isomorphic to $\mathbb{R}$... – 2017-01-07
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    The induced map on $0$th cohomology is determined by the image of path-connected components. How many path-components does each space have and where are they sent under any map? – 2017-01-07
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    $\mathbb{R}$ is path-connected, so 1 component but I don't really see how this helps. I did not see anything about this path-connected component (we only saw de Rham cohomology during the last lecture, so it was quite limited in material) – 2017-01-07
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    The induced map maps $[\omega] \mapsto [f^{\ast} \omega]$, where $[\omega]$ is the homology class of $\omega$ and $f^{\ast}$ is the pullback of $f$. Since these $\omega$ are closed zero-forms (functions form $S^2 \to \mathbb{R}$ with derivative equal to 0), they have to be constant functions and so there can only be one class. The same holds for the image and therefore the induced maps are maps between one point sets, so they have to be the same. Is this correct (more or less)? – 2017-01-07
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    Yep, the composition of any function with a constant function (the constant function applied last) is the same constant function. – 2017-01-07
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    So this proves my question? Well thank you very much! – 2017-01-07