Proof of Theorem 7 (Chapter 5) in Hoffman and Kunze's *Linear Algebra* is unclear

The key observation is the following identity which we state as a lemma.

Lemma: Let $1\leq r \leq n$. Consider the $r$-tuple $(j_1,\dots,j_r)$, $1 \leq j_1,\dots,j_r \leq n$. Let $\sigma$ be a permutation of $\{1,\dots,r\}$. Then, we have $$f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} = (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}.$$

Proof: For $(\alpha_1,\dots,\alpha_r) \in V^r$, we have $$ \begin{align} (f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}})(\alpha_1,\dots,\alpha_r) &= f_{j_{\sigma 1}}(\alpha_1) \cdots f_{j_{\sigma r}}(\alpha_r) \\ &= f_{j_1}(\alpha_{\sigma^{-1} 1}) \cdots f_{j_r}(\alpha_{\sigma^{-1} r}) &&\text{(by rearranging terms)} \\ &= (f_{j_1}\! \otimes \dots \otimes f_{j_r})(\alpha_{\sigma^{-1} 1},\dots,\alpha_{\sigma^{-1} r}) \\ &= (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}(\alpha_1,\dots,\alpha_r). \end{align} $$ Since $(\alpha_1,\dots,\alpha_r)$ was an arbitrary element of $V^r$, we get $$f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} = (f_{j_1}\! \otimes \dots \otimes f_{j_r})_{\sigma^{-1}}.$$ Hence, proved.

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Using this lemma, we show that equation $(5\text{-}39)$ is correct as follows. $$ \begin{align} \pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}) &:= \sum_{\sigma} (\operatorname{sgn} \sigma) (f_{j_1}\! \otimes \dots \otimes f_{j_r})_\sigma\\ &= \sum_\sigma (\operatorname{sgn} \sigma)\ f_{j_{\sigma^{-1} 1}}\! \otimes \dots \otimes f_{j_{\sigma^{-1} r}} &&(\text{by above lemma})\\ &= \sum_\sigma (\operatorname{sgn} \sigma^{-1})\ f_{j_{\sigma^{-1} 1}}\!\otimes \dots \otimes f_{j_{\sigma^{-1} r}} &&(\because \operatorname{sgn} \sigma = \operatorname{sgn} \sigma^{-1}). \end{align} $$ Now, as $\sigma$ runs (once) over all permutations of $\{ 1,\dots,r \}$, so does $\sigma^{-1}$. Therefore, $$ \pi_r(f_{j_1}\! \otimes \dots \otimes f_{j_r}) = \sum_{\tau}(\operatorname{sgn} \tau)\ f_{j_{\tau 1}}\! \otimes \dots \otimes f_{j_{\tau r}} $$ where the sum is taken over all permutations $\tau$ of $\{ 1,\dots,r \}$. But, the right-hand side is precisely $D_J$. Hence, proved.