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Noah Schweber
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Simple proof of a isomorfisman isomorphism theorem

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José Hdz. Stgo.
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I haven't references about a proof of this theorem: Let $p: Y \rightarrow X$ be a fibrefiber bundle. If for all $x \in X$ $p^{-1}(x)$ satisfies $H^{*}(p^{-1}(x)) \simeq \mathbb{R} $$H^{\ast}(p^{-1}(x)) \simeq \mathbb{R} $, then $p$ induces an isomorphism $ H^{*}(X) \simeq H^{*}(Y)$$ H^{\ast}(X) \simeq H^{*}(Y)$. How could I prove this theorem? A spectral sequences argument for the fiber bundle?

I haven't references about a proof of this theorem: Let $p: Y \rightarrow X$ be a fibre bundle. If for all $x \in X$ $p^{-1}(x)$ satisfies $H^{*}(p^{-1}(x)) \simeq \mathbb{R} $, then $p$ induces an isomorphism $ H^{*}(X) \simeq H^{*}(Y)$. How could I prove this theorem? A spectral sequences argument for the fiber bundle?

I haven't references about a proof of this theorem: Let $p: Y \rightarrow X$ be a fiber bundle. If for all $x \in X$ $p^{-1}(x)$ satisfies $H^{\ast}(p^{-1}(x)) \simeq \mathbb{R} $, then $p$ induces an isomorphism $ H^{\ast}(X) \simeq H^{*}(Y)$. How could I prove this theorem? A spectral sequences argument for the fiber bundle?

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Simple proof of a isomorfism theorem

I haven't references about a proof of this theorem: Let $p: Y \rightarrow X$ be a fibre bundle. If for all $x \in X$ $p^{-1}(x)$ satisfies $H^{*}(p^{-1}(x)) \simeq \mathbb{R} $, then $p$ induces an isomorphism $ H^{*}(X) \simeq H^{*}(Y)$. How could I prove this theorem? A spectral sequences argument for the fiber bundle?