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There is a continuous map from $S^n$ to the wedge sum of spheres. I imagine that the induced pull back map on cohomology groups $H^n( S^n \vee S^n) \cong \mathbb{Z} \oplus \mathbb{Z} \rightarrow H^n( S^n ) \cong \mathbb{Z}$ is just addition though I'm not sure. Is there an easy way to show this?

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    $\begingroup$ Look at what it does to the canonical generators. $\endgroup$ – Pedro Tamaroff Apr 8 '16 at 19:10
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    $\begingroup$ I was about to press "enter" when Pedro posted his answer, so I will post it anyhow: You can describe the map $S^n\to S^n\vee S^n$, and then use that to describe what it does to generators on cohomology. $\endgroup$ – John Martin Apr 8 '16 at 19:11
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A map is given explicitly by composing the diagonal map $x \rightarrow (x,x)$ with quotienting by a point. Let $\alpha_1 \in H^n(S^n \vee S^n)$ send the generator of the homology of the the "first" sphere $\sigma_1$ to $1$ and $\alpha_2$ send the generator of the second sphere $\sigma_2$ to $1$. Under the diagonal map the generator of $H_n(S^n)$ gets sent to the sum of generators $\sigma_1+\sigma_2$ so that the pull back of $\alpha_1+\alpha_2$ on the generator of $H_n(S^n)$ gives $f^*(\alpha_1+\alpha_2)(\sigma)=(\alpha_1+\alpha_2)(\sigma_1+\sigma_2)=\alpha_1(\sigma_1)+\alpha_2(\sigma_2)$.

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  • $\begingroup$ Nicely done. It's nice to see that your led through it on your own! +1 $\endgroup$ – John Martin Apr 9 '16 at 1:02

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