--- title: Proof of Leray-Hirsch Theorem --- # Preliminaries **Definition:** Fibre Bundle **Definition:** Homology **Definition:** Cup Product **Definition:** Good Cover **Notation:** Let $R$ be an arbitrary ring, and let $h$ denote the contravariant functor $$ h(\wait; R): \mathbf{Top} &\to\mathbf{Ring} \\ X &\mapsto H_{\text{sing}}^*(X; R) \\ (X \mapsvia{f} Y) &\mapsto (H^*(Y; R) \mapsvia{f^*} H^*(X; R)) $$ which corresponds to taking singular cohomology. > Note that $H^*_{\text{sing}}(X; R)$ is a graded ring with multiplicative structure given by the cup product, and similarly $H^*_{\text{dR}}(X; \RR)$ is graded ring with multiplication induced by the wedge product of forms. # Statement of the Theorem Let ```{=latex} \begin{center} \begin{tikzcd} F \arrow[rr, "i", hook] & & E \arrow[dd, "p", two heads] \\ & & \\ & & B \end{tikzcd} \end{center} ``` be a fibre bundle. Taking cohomology induces maps ```{=latex} \begin{center} \begin{tikzcd} h(F; R) & & h(E; R) \arrow[ll, "i^*"'] \\ & & \\ & & h(B; R) \arrow[uu, "p^*"'] \end{tikzcd} \end{center} ``` Suppose that 1. $h(F; R)$ is a finitely-generated free $R\dash$module in each degree $n$, and 2. For every fiber $F$, there exists a collection of chains $$ C_F \definedas \theset{c_j \mid j \in J}\subseteq h(E; R) \quad \text{for some index set} J $$ such that their restrictions $\theset{i^*(c_j) \mid j \in J} \subseteq h(F; R)$ along $i^*$ yield an $R\dash$basis for $h(F; R)$, i.e. $$ h(F; R) = \spanof_R\left(\theset{i^*(c_j) \mid j\in J}\right) $$ as an $R\dash$module. We can then define the following group action: \begin{align*} h(B; R) &\actson h(E; R) \\ b \actson e &\definedas p^*(b) \smile e ,\end{align*} and as a result we have 1. Both $h(E; R)$ and $h(F; R) \tensor_R h(B; R)$ are modules over the ring $h(B; R)$, 2. Letting $\theset{b_k \mid k\in K} \subseteq h(B; R)$ denote a (not necessarily finite) set of generators, $$ h(B; R) \tensor_R h(F; R) = \spanof_R\theset{b_i \tensor i^*(c_j) \mid k\in K, j\in J} $$ 3. The following map is an isomorphism in the category of $h(B; R)\dash$modules: $$ \varphi: h(B; R) \otimes_R h(F; R) &\to \quad h(E; R) \\ \sum_{k\in K, j \in J} b_k \otimes_R i^*(c_j) \ &\mapsto \sum_{k\in K, j \in J} p^*(b_k) \smile c_j $$ 4. As an $h(B; R)$ module, $$ h_E(R) = \spanof_{h(B; R)}\left( \theset{c_j \mid j\in J} \right) $$ so the cohomology of the total space is given by $h(B; R)$ span of these $c_j$. > Note: this map is not an isomorphism in the category of rings. **Remark:** The assumption that each $C_F$ exists is necessary, and can be guaranteed when $E \cong B \cross F$ is homeomorphic to a product. Letting $p_B: B\cross F \to B$ and $p_F: B\cross F \to F$ be the projections supplied by the universal property of the product, since $h(F; R)$ is a free $R\dash$module, we can take its $R\dash$basis $\theset{f_k \mid j\in K} \subseteq h(F; R)$ and pull it back along $p_F$ to obtain $\theset{p_F^*(f_k)} \subseteq h(F\cross B; R)$. ```{=latex} \begin{center} \begin{tikzcd} F \arrow[rr, "i", dotted, bend left] & & F \cross B \arrow[dd, "p_B"] \arrow[ll, "p_F"] & && & h(F; R) \arrow[rr, "p_F^*"'] & & h(F\cross B; R) \arrow[dd, "p_B^*"] \arrow[ll, "i^*"', dotted, bend right] \\ & && & & & & & \\ & & B& && && & h(B;R) \end{tikzcd} \end{center} ``` Then we can note that for a product, $p_F \circ i = \id_{F\cross B}$, and so $i^* \circ p_F^* = \id_F$. Thus defining $c_k \definedas p^*_F(f_k)$ satisfies condition (2), i.e. $i^* (c_k) \definedas (i^* p_F^*)(f_k) = f_k$ is an $R\dash$basis for $h(F; R)$. **Corollary:** Let $h(X) = H_{dR}^*(X; \RR)$ denote taking deRham cohomology. If 1. $F \to E \to M$ is a fibre bundle of smooth manifolds where $M$ has a good cover, 2. There exist a set of global forms $\theset{\omega_j} \subseteq E$ such that $h(E)=\spanof_\RR \theset{\omega_j \mid j\in J}$, and 3. For each fiber $F$, the collection of restrictions $\theset{\restrictionof{\omega_j}{F} \mid j\in J}$ freely generated $h(F)$, then $h(E)$ is a free $h(M)\dash$module and $$ h(E) = \spanof_\RR \theset{\omega_j \mid j\in J} \tensor h(M) \cong h(F) \tensor h(M). $$ # Proof We'll prove the special case given in the corollary. Given a fibre bundle $F \to E \to M$, there are projections $p_F$ and $p_M$ which induce maps in the deRham complex, ```{=latex} \begin{center} \begin{tikzcd} F & & E \arrow[dd, "p_B"] \arrow[ll, "p_F"] & && & \Omega^*(F) \arrow[rr, "p_F^*"'] & & \Omega^*(E) \arrow[dd, "p_B^*"] \\ & & & & & & & & \\ & & M & & & & & & \Omega^*(M) \end{tikzcd} \end{center} ``` This allows us to define a map on forms: \begin{align*} \phi: \Omega^*(M) \tensor_\RR \Omega^*(F) &\longrightarrow \Omega^*(E) \\ \omega \tensor_\RR \phi \quad &\mapsto \quad p_B^*(\omega) \wedge p_F^*(\phi) ,\end{align*} which amounts to pulling the forms back and then wedging them. **Claim:** $\phi$ induces a map on the deRham cohomology $h^*$. By the claim, we obtain a map \begin{align*} \tilde \phi: h(M) \tensor_\RR h(F) &\longrightarrow h(E)\\ [\omega] \tensor_\RR [\phi] \quad &\mapsto \quad [p_B^*(\omega) \wedge p_F^*(\phi)] .\end{align*} **Claim:** $\phi$ is an isomorphism of graded modules (???). Proof: We will induct on the cardinality of the good cover $\mathcal U$ of $M$. For the base case, suppose $\# \mathcal{U} = 2$, so $M = U \union V$. Noting that here $h^*(X) \definedas H^*_\text{dR}(X; \RR)$ is a graded ring, we can identify $$ h(M) \tensor_\RR h(F) = \bigoplus_{i+j = n} h^i(M) \tensor_\RR h^{j}(F) $$ and so the $k$th graded piece is given by $$ \left( h(M) \tensor_\RR h(F) \right)^k = \bigoplus_{i+j = k} h^i(M) \tensor_\RR h^{j}(F). $$ Similarly, $h(E) = \bigoplus_{i=0}^n h^i(E)$, and thus $\phi$ is an isomorphism iff it is an isomorphism between $k$th graded pieces for every $k$. So it suffices to show that the maps \begin{align*} \tilde \phi_k : \bigoplus_{i+j = k} h^i(M) \tensor_\RR h^{j}(F) \to h^k(E) \\ .\end{align*} are isomorphisms for every $0 \leq k \leq n$. To this end, fix an arbitrary $0\leq k \leq n$ and consider the following diagram: ```{=latex} \vspace{4em} \begin{figure}[!htbp] \begin{center} \begin{tikzcd}[transform canvas={scale=0.70}, column sep=1.2em] 0 & \bigoplus_{j=0}^{k+1} h^{j}(U\cap V) \tensor_R h^{n-k}(F) \arrow[l] \arrow[dd, "\tilde\phi_{k+1}"]& \bigoplus_{j=0}^k h^j(U\cup V) \tensor_R h^{n-k}(F) \arrow[l] \arrow[dd, "\tilde \phi_k"]& \bigoplus_{j=0}^k (h^j(U) \tensor h^{n-k}(F)) \oplus (h^j(V)\tensor h^{n-k}(F)) \arrow[l] \arrow[dd, "\tilde \phi_k \oplus \tilde \phi_k"]& \bigoplus_{j=0}^k h^j(U\cap V) \tensor_R h^{n-k}(F) \arrow[l] \arrow[dd, "\tilde\phi_k"]& 0 \arrow[l] \\ \\ 0 & h^{k+1}(U\cap V \cross F) \arrow[l] & \arrow[l] h^k(E) & h^k(U \cross F) \bigoplus h^k(V\cross F) \arrow[l] & h^k(U\cap V \cross F) \arrow[l] & 0 \arrow[l] \end{tikzcd} \vspace{4em} \end{center} \caption{Main Diagram} \end{figure} ``` **Claim:** - The rows in this diagram are exact, - The diagram commutes, and - The 1st, 3rd, and 4th maps are isomorphisms. If this is the case, the 5 lemma can be applied and this will imply that the 2nd map \begin{align*} \tilde \phi_k:~\bigoplus_{j=0}^{k} h^{j}(U\cup V) \tensor_R h^{n-k}(F) &\longrightarrow h^{k}(E) \end{align*} is an isomorphism. Since $k$ was arbitrary, $\tilde \phi_k$ will be an isomorphism for every $k$, which is precisely what we want to show. ## The Top Row is Exact By Mayer-Vietoris, there is a long exact sequence: ```{=latex} \begin{tikzcd} 0 & & h^n(U\cup V) \arrow[ll] & & h^n(U) \oplus h^n(V) \arrow[ll] & & h^n(U\cap V) \arrow[ll] & & \\ & & & & & & & & \\ & & \arrow[rrrruu, "\delta"] h^{n-1}(U\cup V) & & h^{n-1}(U) \oplus h^{n-1}(V) \arrow[ll] & & h^{n-1}(U\cap V) \arrow[ll] & & \\ & & & & & & & & \\ & & \arrow[rrrruu, dashed, "\delta"] h^k(U\cup V) & & h^k(U) \oplus h^k(V) \arrow[ll] & & h^k(U\cap V) \arrow[ll] & & \\ & & & & & & & & \\ & &\arrow[rrrruu, dashed, "\delta"] h^0(U\cup V) & & h^0(U) \oplus h^0(V) \arrow[ll] & & h^0(U\cap V) \arrow[ll]& & \arrow[ll] 0 \end{tikzcd} ``` where $\delta$ is the connecting map supplied by the Snake Lemma. Since $h^*(F)$ was assumed to be a free $R\dash$module, the functor $(\wait) \tensor_R h^j(F)$ is exact for any $j$. \newpage Fixing $j$ for the moment, we note that applying $(\wait) \tensor h^j(F)$ to the above sequence yields a new long exact sequence: ```{=latex} \vspace{8em} \begin{center} \begin{tikzcd}[transform canvas={scale=0.90}, column sep=0.6em] 0 & & \arrow[ll] h^n(U\cup V) \tensor_R h^j(F) & & \arrow[ll] h^n(U) \tensor_R h^j(F) \oplus h^n(V) \tensor_R h^j(F) & & \arrow[ll] h^n(U\cap V) \tensor_R h^j(F) & & \\ & & & & & & & & \\ & & \arrow[rrrruu, "\delta \tensor \id_{h^j(F)}"] h^{n-1}(U\cup V) \tensor_R h^j(F) & & h^{n-1}(U) \tensor_R h^j(F) \oplus h^{n-1}(V) \tensor_R h^j(F) \arrow[ll] & & h^{n-1}(U\cap V) \tensor_R h^j(F) \arrow[ll] & & \\ & & & & & & & & \\ & &\arrow[rrrruu, dashed, "\delta \tensor \id_{h^j(F)}"] h^k(U\cup V) \tensor_R h^j(F) & & h^k(U) \tensor_R h^j(F) \oplus h^k(V) \tensor_R h^j(F) \arrow[ll] & & h^k(U\cap V) \tensor_R h^j(F) \arrow[ll] & & \\ & & & & & & & & \\ & &\arrow[rrrruu, dashed, "\delta \tensor \id_{h^j(F)}"] h^0(U\cup V) \tensor_R h^j(F) & & h^0(U) \tensor_R h^j(F) \oplus h^0(V) \tensor_R h^j(F) \arrow[ll] & & h^0(U\cap V) \tensor_R h^j(F) \arrow[ll] & & \arrow[ll] 0 \ \end{tikzcd} \vspace{8em} \end{center} ```` where the tensor product has been distributed across the direct sums in the middle column. Consider doing this for every $j$; we then obtain a collection of long exact sequences ```{=latex} \vspace{8em} \begin{center} \begin{tikzcd}[transform canvas={scale=0.75}, column sep=0.5em] j=k+1& && && & 0 \arrow[rr]& & h^0(U\union V) \tensor h^{k+1}(F) \arrow[rr] & & \cdots \\ % j=k& 0 \arrow[rr] & & h^0(U \union V) \tensor h^k(F) \arrow[rr] & & h^0(U) \tensor h^k(F) \oplus h^0(V) \tensor h^k(F) \arrow[r] & h^0(U \intersect V) \tensor h^k(F) \arrow[rr, "\delta"] & & h^1(U\union V) \tensor h^{k}(F) \arrow[rr] & & \cdots \\ % j=k-1:& \cdots \arrow[rr, "\delta"] & & h^1(U \union V) \tensor h^{k-1}(F) \arrow[rr] & & h^1(U) \tensor h^{k-1}(F) \oplus h^1(V) \tensor h^{k-1}(F) \arrow[r] & h^1(U \intersect V) \tensor h^{k-1}(F) \arrow[rr, "\delta"] & & h^2(U\union V) \tensor h^{k-1}(F) \arrow[rr] & & \cdots \\ % j=k-2:& \cdots \arrow[rr, "\delta"] & & h^2(U \union V) \tensor h^{k-2}(F) \arrow[rr] & & h^2(U) \tensor h^{k-2}(F) \oplus h^2(V) \tensor h^{k-2}(F) \arrow[r] & h^2(U \intersect V) \tensor h^{k-2}(F) \arrow[rr, "\delta"] & & h^2(U\union V) \tensor h^{k-2}(F) \arrow[rr] & & \cdots \\ % \vdots & & & \vdots & & \vdots & \vdots & & \vdots & & \vdots \\ % j=0: & \cdots \arrow[rr, "\delta"] && h^k(U \union V) \tensor h^0(F) \arrow[rr] && h^k(U) \tensor h^0(F) \oplus h^k(V) \tensor h^0(F) \arrow[r] & h^k(U \intersect V) \tensor h^0(F) \arrow[rr, "\delta"] & & h^{k+1}(U\union V) \tensor h^{0}(F) \arrow[rr] && \cdots \\ \end{tikzcd} \end{center} \vspace{8em} ``` Then summing along the columns will preserve exactness in each each degree. Moreover, taking the direct sum down the first, second, third, and fourth columns respectively yields \begin{align*} \text{Column 1}: & C_1 \definedas \bigoplus_{j=0}^k h^j(U \union V) \tensor h^{k-j}(F) \\ \\ \text{Column 2}: & C_2 \definedas \bigoplus_{j=0}^k h^j(U) \tensor h^{k-j}(F) \oplus h^j(V) \tensor h^{k-j}(F) \\ \\ \text{Column 3}: & C_3\definedas \bigoplus_{j=0}^k h^j(U \intersect V)\tensor h^{k-j}(F) \\ \\ \text{Column 4}: & C_4 \definedas \bigoplus_{j=0}^{k+1} h^j(U \union V) \tensor h^{k-j}(F) \\ \\ ,\end{align*} and the exactness of the sequence $C_1 \to C_2 \to C_3 \to C_4$ is precisely the exactness of the top row in figure (1). $$ ## The Bottom Row is Exact