Definition of a Schur Functor
Eisenbud (p. 591) defines the schur functor to be a composite of two maps one of which is the map $\Delta^d: \wedge^d(M) \rightarrow T_d(M) = M^{\otimes d}$ obtained from the diagonalization map $\Delta: \wedge(M) \rightarrow \wedge(M \oplus M) = \wedge(M) \otimes \wedge(M)$ defined by $m \mapsto (m,m) \mapsto m \otimes 1 +1 \otimes m$.
I’m having a hard time understanding this map. I’m guessing this is the embedding map $\psi: \wedge(M) \rightarrow \otimes(M)$ defined by $m_1 \wedge m_2 \wedge \cdots \wedge m_n \rightarrow \sum_{\sigma \in S_n} (\operatorname{sgn} \sigma)m_{\sigma(1)} \otimes \cdots \otimes m_{\sigma(n)}$ since it seems to make sense with the examples Eisenbud gives. However, I’m not sure how this map is derived from the diagonal map.
For example,
$\psi(m \wedge n) = m \otimes n - n \otimes m$
but
$\Delta(m \wedge n) = (1 \otimes m + m \otimes 1) \wedge (1 \otimes n + n \otimes 1) = 1 \otimes (m \wedge n) + n \otimes m + m \otimes n + (m \wedge n) \otimes 1.$
Now $m \otimes n + n \otimes m \in M \otimes M$ but there is no minus sign...
