[proofplan] The tautological bundle $S$ sits as a subbundle of the product bundle $\underline{\mathbb C}^n$ over the Grassmannian, and the quotient is precisely $Q$. Applying the Whitney product formula to this short exact sequence gives $c(S)c(Q)=c(\underline{\mathbb C}^n)$. The product bundle has total Chern class $1$, so the desired multiplicative relation follows. Expanding the product degree by degree gives the equivalent relations among the individual Chern classes. [/proofplan]

[step:Identify the tautological short exact sequence] Let $X:=\operatorname{Gr}(k,n)$, let $\pi_S:S\to X$ denote the tautological bundle, let $\pi_{\mathrm{triv}}:\underline{\mathbb C}^n=X\times\mathbb C^n\to X$ denote the product bundle, and let $\pi_Q:Q\to X$ denote the quotient bundle. For each point $V\in X$, the inclusion of the fiber $S_V=V$ into $\mathbb C^n$ gives a fiberwise injective bundle map \begin{align*} \iota:S\longrightarrow \underline{\mathbb C}^n. \end{align*} By definition of the quotient bundle, $Q$ is the cokernel bundle of this inclusion, with fiber \begin{align*} Q_V=\mathbb C^n/V. \end{align*} Hence there is a short exact sequence of complex vector bundles over $X$: \begin{align*} 0\longrightarrow S\stackrel{\iota}{\longrightarrow}\underline{\mathbb C}^n\longrightarrow Q\longrightarrow 0. \end{align*} [/step]

[step:Apply the Whitney product formula to the tautological exact sequence]We use the [Whitney product formula for Chern classes](/theorems/7052) [citetheorem:9797]: if \begin{align*} 0\longrightarrow E'\longrightarrow E\longrightarrow E''\longrightarrow 0 \end{align*} is a short exact sequence of complex vector bundles over a space $Y$, then \begin{align*} c(E)=c(E')c(E'') \end{align*} in $H^*(Y;\mathbb Z)$. The sequence from the previous step is a short exact sequence of complex vector bundles over $X$, so the formula applies with $E'=S$, $E=\underline{\mathbb C}^n$, and $E''=Q$. Therefore \begin{align*} c(\underline{\mathbb C}^n)=c(S)c(Q). \end{align*}[/step]

[guided]The purpose of this step is to convert the geometric exact sequence into an identity in cohomology. The relevant input is the Whitney product formula for Chern classes [citetheorem:9797], which says that for every short exact sequence of complex vector bundles \begin{align*} 0\longrightarrow E'\longrightarrow E\longrightarrow E''\longrightarrow 0 \end{align*} over a space $Y$, the total Chern classes satisfy \begin{align*} c(E)=c(E')c(E'') \end{align*} in the [cohomology ring](/theorems/2271) $H^*(Y;\mathbb Z)$. We verify the hypotheses in the present situation. The base space is $Y=X=\operatorname{Gr}(k,n)$. The bundle $S\to X$ is a complex vector bundle because its fiber over $V\in X$ is the complex [vector space](/page/Vector%20Space) $V$. The bundle $\underline{\mathbb C}^n=X\times\mathbb C^n\to X$ is a complex vector bundle. The bundle $Q\to X$ is the quotient complex vector bundle with fiber $Q_V=\mathbb C^n/V$. The maps \begin{align*} 0\longrightarrow S\stackrel{\iota}{\longrightarrow}\underline{\mathbb C}^n\longrightarrow Q\longrightarrow 0 \end{align*} are complex-linear on each fiber, and the sequence is exact on each fiber because $S_V=V\subset \mathbb C^n$ and $Q_V=\mathbb C^n/V$. Thus the Whitney product formula applies with $E'=S$, $E=\underline{\mathbb C}^n$, and $E''=Q$. It gives \begin{align*} c(\underline{\mathbb C}^n)=c(S)c(Q) \end{align*} in $H^*(X;\mathbb Z)$.[/guided]

[step:Use the normalization of Chern classes on the product bundle] The product bundle $\underline{\mathbb C}^n$ has total Chern class \begin{align*} c(\underline{\mathbb C}^n)=1. \end{align*} This is the normalization property of Chern classes for product complex vector bundles, equivalently the vanishing of all positive-degree Chern classes of such a bundle. Substituting this into the identity obtained above gives \begin{align*} c(S)c(Q)=1 \end{align*} in $H^*(X;\mathbb Z)$. [/step]

[step:Read off the positive-degree relations] Write \begin{align*} c(S)=1+s_1+\cdots+s_k \end{align*} with $s_i:=c_i(S)\in H^{2i}(X;\mathbb Z)$, and write \begin{align*} c(Q)=1+q_1+\cdots+q_{n-k} \end{align*} with $q_j:=c_j(Q)\in H^{2j}(X;\mathbb Z)$. Define $s_0:=1$, $q_0:=1$, set $s_i:=0$ for $i>k$, and set $q_j:=0$ for $j>n-k$. The degree-$2d$ component of the product $c(S)c(Q)$ is then \begin{align*} \sum_{i+j=d}s_iq_j. \end{align*} Since $c(S)c(Q)=1$, all components of positive cohomological degree vanish. Therefore, for every integer $d\ge 1$, \begin{align*} \sum_{i+j=d}s_iq_j=0 \end{align*} in $H^{2d}(X;\mathbb Z)$. This is exactly the stated equivalent form of the Chern relation. [/step]