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1.

For every $k$-vector space $V$, and every $n\in\mathbb N$, the symmetric group $S_n$ acts on the tensor power $V^{\otimes n}$ by permuting the tensorands:tensorands:

$\sigma\left(v_1\otimes v_2\otimes ...\otimes v_n\right) = v_{\sigma^{-1}\left(1\right)}\otimes v_{\sigma^{-1}\left(2\right)}\otimes ...\otimes v_{\sigma^{-1}\left(n\right)}$ for every $\sigma \in S_n$ and $v_1,v_2,...,v_n\in V$.

Let $V^{\otimes n}_{\mathrm{symm}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts trivially. Let $V^{\otimes n}_{\mathrm{alt}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts by the sign representation.representation.

For every $k$-linear map $f:V\to W$ between two vector spaces, its $n$-th tensor power $f^{\otimes n}:V^{\otimes n}\to W^{\otimes n}$ restricts to a $k$-linear map $f^{\otimes n}_{\mathrm{symm}}:V^{\otimes n}_{\mathrm{symm}}\to W^{\otimes n}_{\mathrm{symm}}$ and a $k$-linear map $f^{\otimes n}_{\mathrm{alt}}:V^{\otimes n}_{\mathrm{alt}}\to W^{\otimes n}_{\mathrm{alt}}$, because the map $f$ commutes with the action of $S_n$ (while $f$ transforms the tensorands, the action of $S_n$ permutes the tensorands). Thus, $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ and $V\mapsto V^{\otimes n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$ are functors.

2.

If $k$ has characteristic $0$, then $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ is isomorphic to $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ as functors, and $V\mapsto V^{\otimes n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$ is isomorphic to $V\mapsto \wedge^n\left(V\right);\ f\mapsto \wedge^n\left(f\right)$ as functors. The isomorphisms are given, e. g., in Crawley-Boevey, Lectures on representation theory and invariant theory, §6, Lemma 1 and 3.

Of course, multiplying such an isomorphism by a scalar $\neq 0$ yields another isomorphism. This is all the freedom we have: any two isomorphisms between the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ and the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ are equal up to scalar, and similarly for the other pair of functors.

To prove this, we let $P$ be an isomorphism from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$. Let $\lambda\in k$ be defined by $P_k\left(1\otimes 1\otimes ...\otimes 1\right)=\lambda 1\cdot 1\cdot ...\cdot 1$, where $P_k$ is the isomorphism $P$ at the object $V=k$, and $\cdot$ denotes the multiplication in the symmetric algebra (because it is commutative).

Since $k$ has characteristic $0$, the space $V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes ...\otimes v\otimes\dots\otimes v$ for $v\in V$. (This is Lemma 7 from Crawley-Boevey's above-mentioned text, sent back to $V^{\otimes n}_{\mathrm{symm}}$ from $\mathrm{S}^n\left(V\right)$.) We are now going to prove that $P_V\left(v\otimes v\otimes ...\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot ...\cdot \ldots\cdot v$ for every $v\in V$.V$. In order to show this, let$f:k\to V$be a vector space homomorphism given by$f\left(1\right)=v$. The functoriality of$P$now yields $P_V\left(f^{\otimes n}_{\mathrm{symm}}\left(1\otimes 1\otimes ...\otimes 1\otimes\ldots\otimes 1\right)\right)=\left(\mathrm{S}^n\left(f\right)\right)\left(P_k\left(1\cdot 1\cdot ...\cdot 1\cdot\ldots\cdot 1\right)\right)$. This rewrites as $P_V\left(v\otimes v\otimes ...\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot ...\cdot \ldots\cdot v$, and we are done. This yields (since the space $ V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors$v\otimes v\otimes ...\otimes v\otimes\ldots\otimes v$for$v\in V$) that the map $P_V$ is just the canonical projection from $ V^{\otimes n}_{\mathrm{symm}}$ to$\mathrm{S}^n\left(V\right)$, multiplied with the scalar$\lambda$. Since$\lambda$does not depend on$V$, this shows us that our isomorphism$P$from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ is the projection isomorphism times$\lambda$. In other words, all the freedom we have to choose this isomorphism is the freedom of choosing the scalar factor to multiply with. The same argument works for the other pair of functors. 3 Rollback to Revision 1 For every$k$-linear map$f:V\to W$between two vector spaces, its$n$-th tensor power$f^{\otimes n}:V^{\otimes n}\to W^{\otimes n}$restricts to a$k$-linear map$f^{\otimes n}{\mathrm{symm}}:V^{\otimes n}{\mathrm{symm}}\to n}_{\mathrm{symm}}:V^{\otimes n}_{\mathrm{symm}}\to W^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ and a $k$-linear map $f^{\otimes n}{\mathrm{alt}}:V^{\otimes n}{\mathrm{alt}}\to n}_{\mathrm{alt}}:V^{\otimes n}_{\mathrm{alt}}\to W^{\otimes n}{\mathrm{alt}}$, n}_{\mathrm{alt}}$, because the map$f$commutes with the action of$S_n$(while$f$transforms the tensorands, the action of$S_n$permutes the tensorands). Thus,$V\mapsto V^{\otimes n}{\mathrm{symm}};\ n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ and $V\mapsto V^{\otimes n}{\mathrm{alt}};\ n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}{\mathrm{alt}}$ n}_{\mathrm{alt}}$are functors. If$k$has characteristic$0$, then$V\mapsto V^{\otimes n}{\mathrm{symm}};\ n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ is isomorphic to $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ as functors, and $V\mapsto V^{\otimes n}{\mathrm{alt}};\ n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}{\mathrm{alt}}$ n}_{\mathrm{alt}}$is isomorphic to$V\mapsto \wedge^n\left(V\right);\ f\mapsto \wedge^n\left(f\right)$as functors. The isomorphisms are given, e. g., in Crawley-Boevey, Lectures on representation theory and invariant theory, §6, Lemma 1 and 3. Of course, multiplying such an isomorphism by a scalar$\neq 0$yields another isomorphism. This is all the freedom we have: any two isomorphisms between the functor$V\mapsto V^{\otimes n}{\mathrm{symm}};\ n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ and the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ are equal up to scalar, and similarly for the other pair of functors.

To prove this, we let $P$ be an isomorphism from the functor $V\mapsto V^{\otimes n}{\mathrm{symm}};\ n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}{\mathrm{symm}}$ n}_{\mathrm{symm}}$to the functor$V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$. Let$\lambda\in k$be defined by$P_k\left(1\otimes 1\otimes ...\otimes 1\right)=\lambda 1\cdot 1\cdot ...\cdot 1$, where$P_k$is the isomorphism$P$at the object$V=k$, and$\cdot$denotes the multiplication in the symmetric algebra (because it is commutative). Since$k$has characteristic$0$, the space$ V^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes ...\otimes v$ for $v\in V$. (This is Lemma 7 from Crawley-Boevey's above-mentioned text, sent back to $V^{\otimes n}{\mathrm{symm}}$ n}_{\mathrm{symm}}$from$\mathrm{S}^n\left(V\right)$.) We are now going to prove that$P_V\left(v\otimes v\otimes ...\otimes v\right)=\lambda v\cdot v\cdot ...\cdot v$for every$v\in V$. This yields (since the space$ V^{\otimes n}_{\mathrm{symm}}$is generated by the tensors$v\otimes v\otimes ...\otimes v$for$v\in V$) that the map$P_V$is just the canonical projection from$ V^{\otimes n}{\mathrm{symm}}$n}_{\mathrm{symm}}$ to $\mathrm{S}^n\left(V\right)$, multiplied with the scalar $\lambda$. Since $\lambda$ does not depend on $V$, this shows us that our isomorphism $P$ from the functor $V\mapsto V^{\otimes n}{\mathrm{symm}};\ n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ is the projection isomorphism times $\lambda$. In other words, all the freedom we have to choose this isomorphism is the freedom of choosing the scalar factor to multiply with. The same argument works for the other pair of functors.`

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For every $k$-linear map $f:V\to W$ between two vector spaces, its $n$-th tensor power $f^{\otimes n}:V^{\otimes n}\to W^{\otimes n}$ restricts to a $k$-linear map $f^{\otimes n}_{\mathrm{symm}}:V^{\otimes n}_{\mathrm{symm}}\to n}{\mathrm{symm}}:V^{\otimes n}{\mathrm{symm}}\to W^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$and a$k$-linear map$f^{\otimes n}_{\mathrm{alt}}:V^{\otimes n}_{\mathrm{alt}}\to n}{\mathrm{alt}}:V^{\otimes n}{\mathrm{alt}}\to W^{\otimes n}_{\mathrm{alt}}$, n}{\mathrm{alt}}$, because the map $f$ commutes with the action of $S_n$ (while $f$ transforms the tensorands, the action of $S_n$ permutes the tensorands). Thus, $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ n}{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$and$V\mapsto V^{\otimes n}_{\mathrm{alt}};\ n}{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$n}{\mathrm{alt}}$ are functors.

If $k$ has characteristic $0$, then $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ n}{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$is isomorphic to$V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$as functors, and$V\mapsto V^{\otimes n}_{\mathrm{alt}};\ n}{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$n}{\mathrm{alt}}$ is isomorphic to $V\mapsto \wedge^n\left(V\right);\ f\mapsto \wedge^n\left(f\right)$ as functors. The isomorphisms are given, e. g., in Crawley-Boevey, Lectures on representation theory and invariant theory, §6, Lemma 1 and 3.

Of course, multiplying such an isomorphism by a scalar $\neq 0$ yields another isomorphism. This is all the freedom we have: any two isomorphisms between the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ n}{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$and the functor$V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$are equal up to scalar, and similarly for the other pair of functors. To prove this, we let$P$be an isomorphism from the functor$V\mapsto V^{\otimes n}_{\mathrm{symm}};\ n}{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$n}{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$. Let $\lambda\in k$ be defined by $P_k\left(1\otimes 1\otimes ...\otimes 1\right)=\lambda 1\cdot 1\cdot ...\cdot 1$, where $P_k$ is the isomorphism $P$ at the object $V=k$, and $\cdot$ denotes the multiplication in the symmetric algebra (because it is commutative).

Since $k$ has characteristic $0$, the space $V^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$is generated by the tensors$v\otimes v\otimes ...\otimes v$for$v\in V$. (This is Lemma 7 from Crawley-Boevey's above-mentioned text, sent back to$V^{\otimes n}_{\mathrm{symm}}$n}{\mathrm{symm}}$ from $\mathrm{S}^n\left(V\right)$.) We are now going to prove that $P_V\left(v\otimes v\otimes ...\otimes v\right)=\lambda v\cdot v\cdot ...\cdot v$ for every $v\in V$.

This yields (since the space $V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes ...\otimes v$ for $v\in V$) that the map $P_V$ is just the canonical projection from $V^{\otimes n}_{\mathrm{symm}}$ n}{\mathrm{symm}}$to$\mathrm{S}^n\left(V\right)$, multiplied with the scalar$\lambda$. Since$\lambda$does not depend on$V$, this shows us that our isomorphism$P$from the functor$V\mapsto V^{\otimes n}_{\mathrm{symm}};\ n}{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$to the functor$V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$is the projection isomorphism times$\lambda\$. In other words, all the freedom we have to choose this isomorphism is the freedom of choosing the scalar factor to multiply with. The same argument works for the other pair of functors.

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