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claim for arbitrary homology theories retracted
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algori
algori
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Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary [and the argument works for any cohomology theory].

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary [and the argument works for any cohomology theory].

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary.

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algori
algori
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Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong H^*(D^n\times F,S^{n-1}\times F)$ by excision$H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary [and the argument works for any cohomology theory].

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong H^*(D^n\times F,S^{n-1}\times F)$ by excision and so $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary.

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]

So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary [and the argument works for any cohomology theory].

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algori
algori
  • 23.5k
  • 3
  • 100
  • 152

Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.

Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong H^*(D^n\times F,S^{n-1}\times F)$ by excision and so $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary.