What is $(V^*)^{\otimes l} \stackrel{f}{\rightarrow} S^l( (V^*)^{\oplus l} )\cong S^l( (V^{\oplus l})^* ) \stackrel{g}{\rightarrow} S^l(V^*)$? Let's work over an algebraically closed field and let $V$ be a finite dimensional vector space. Then is there a natural chain of maps 
$$
(V^*)^{\otimes l} \stackrel{f}{\rightarrow} S^l( (V^*)^{\oplus l}  )\cong S^l( (V^{\oplus l})^*  )
\stackrel{g}{\rightarrow} S^l(V^*),
$$
where $S^l(W)$ is the symmetric power of a finite dimensional vector space $W$ and $W^*$ means the set of linear functionals on $W$? 
Note that $g$ is supposed to be induced by the natural diagonal embedding $V\hookrightarrow V^{\oplus l}$ and note $S^l( (V^*)^{\oplus l})=\oplus_{p_1+\ldots +p_l=l} \otimes_{i=1}^l S^{p_i}(V^*)$.


When $l=2$, the above simplifies as: 
    $$ 
(V^*)^{\otimes 2}\stackrel{f}{\rightarrow} S^0(V^*)\otimes S^2(V^*)\oplus S^1(V^*)\otimes S^1(V^*)\oplus S^2(V^*)\otimes S^0(V^*)\stackrel{g}{\rightarrow}S^2(V^*). 
$$ 
    Then does $f$ send $f_1\otimes f_2\mapsto f_1 f_2 \oplus 
(\sum_{\sigma\in S_2} f_{\sigma(1)}\otimes f_{\sigma(2)}) 
\oplus f_1 f_2$, and then does $g$ send 
    $f_1 f_2 \oplus f_1\otimes f_2 \oplus f_1 f_2\mapsto f_1f_2$? 
    If the sequence of maps that I wrote down is correct, then what is the point of embedding $(V^*)^{\otimes l}$ through $f$ into the symmetric power of a larger vector space? 


 A: Let us discuss the $l=2$ case. The point is that
$$V=S^1(V):=T(V)/(I_V),$$
where $I_V$ is the two sided ideal generated by $v_1\otimes v_2-v_2\otimes v_1$, for all $v_1, v_2\in V$. Analogously, we have the embedding (or the definition)
$$V^{*}\otimes V^{*}=S^1(V^{*}\otimes V^{*}).$$ 
If we want to embed $V^{*}\otimes V^{*}$ into the weight $2$ component $S^2(\cdot)$ of a symmetric algebra, we need to "double the dimension". Let us do it. First of all, in what follows we use the injections $W\hookrightarrow W\oplus W$ on the first resp. second factor given by $w\mapsto w\oplus 1$, resp. $w\mapsto 1\oplus w$. The morphism 
$$V^{*}\otimes V^{*}\stackrel{f}{\rightarrow} S^2(V^{*}\oplus V^{*})$$
is given by the composition 
$$V^{*}\otimes V^{*}\rightarrow (V^{*}\oplus V^{*})\otimes (V^{*}\oplus V^{*})\twoheadrightarrow S^2(V^{*}\oplus V^{*})=((V^{*}\oplus V^{*})\otimes (V^{*}\oplus V^{*}))/I_{V^{*}\oplus V^{*}} $$ 
with
$$\epsilon_1\otimes \epsilon_2 \mapsto (\epsilon_1\oplus 1)\otimes (1\oplus \epsilon_2) + I_{V^{*}\oplus V^{*}},$$
for any basis $\{\epsilon_i\}$ of $V^{*}$. The isomorphism 
$$S^2(V^{*}\oplus V^{*})\stackrel{\mathcal I}{\rightarrow} S^2((V\oplus V)^{*})=
((V\oplus V)^{*}\otimes (V\oplus V)^{*})/I_{(V\oplus V)^{*}}$$
is given by (on the image of $f$)
$$(\epsilon_1\oplus 1)\otimes (1\oplus \epsilon_2) + I_{V^{*}\oplus V^{*}}\mapsto 
(e_1\oplus 1)\otimes (1\oplus e_2) + I_{(V\oplus V)^{*}},
 $$
denoting by $\{e_i\}$ the dual basis in $V$, i.e.  $\epsilon_i(e_j)=\delta_{ij}$.
We arrive at the morphism $g$
$$S^2((V\oplus V)^{*})\stackrel{g}{\rightarrow} S^2(V^{*})=
(V^{*}\otimes V^{*})/I_{V^{*}}$$
given by (on the image of $\mathcal I\circ f$)
$$(e_1\oplus 1)\otimes (1\oplus e_2) + I_{(V\oplus V)^{*}}\mapsto 
(\epsilon_1\otimes \epsilon_2) + I_{V^{*}},$$
and we are done.
