Hi I was working through symmetric tensors (technically in a differential manifold, but we defined it abstractly). We defined them first in the usual way: Consider $A$ an integral domain and $\mathbb{V}$ an $A$- module. $T^k(\mathbb{V}) = \underset{ k-times}{ \mathbb{V} \underset{A}{\otimes} ... \underset{A}{\otimes} \mathbb{V}} $.
We have a right side action $T^k(\mathbb{V}) \curvearrowleft \mathbb{S}_k$ given by: if $\tau = v_1 \otimes ... \otimes v_k$ then $$\tau^\delta := v_{\delta(1)} \otimes ... \otimes v_{\delta(k)}$$ For finite sums with coefficients in $A$ extend by linearity: $$\left( \underset{ i=1}{\overset{m}{\sum}}\tau_i\right)^ \delta := \underset{ i=1}{\overset{m}{\sum}}\tau_i ^ \delta $$
Then consider the $A$-submodule $\Sigma_n := \langle \tau^\delta - \tau : \tau \in T^k(\mathbb{V}), \delta \in \mathbb{S}_n \rangle $, that is, the n-submodule generated by elements $\tau^\delta - \tau$ where $\tau$ is an n-tensor and $\delta$ a rearrangement of $\{1, ..., n\}$. Then the symmetric tensors are just $ S^k := T^k(\mathbb{V}) / \Sigma_k $.
The alternative way to define them is to use \begin{align*}\sigma : T^k(\mathbb{V}) & \longrightarrow T^k(\mathbb{V}) \\ \tau & \longrightarrow \sum_{ \delta \in \mathbb{S}_k} \frac{1}{n!} \ \tau^\delta \end{align*}
$\sigma$ is called the "symmetrization" operator. And then define $S^{\prime^{\ \ k}} := Im \ \sigma \subseteq T^k(\mathbb{V})$
You have the composition of morphisms $\pi \circ i $
$S^{\prime^{\ \ k}}$ $\overset{i}{\longrightarrow}$ $T^k(\mathbb{V})$ $\overset{\pi}{\longrightarrow}$ $S^k$
where $i$ is the inclusion of $Im\ \sigma$ and $\pi$ the projection.
THEOREM
$ \pi \circ i : S^{\prime^{\ \ k}} = Im\ \sigma \rightarrow S^k$ is an isomorphism
QUESTION
My problem is to show that is a monomorphism. You have to show that if $\pi ( \sigma(t) ) = 0$, that is, if $\sigma(t) \in \Sigma_n = \langle \tau^\delta - \tau : \tau \in T^k(\mathbb{V}), \delta \in \mathbb{S}_n \rangle$ then $\sigma(t) = 0$. In other words $Im \ \sigma \bigcap \Sigma_n = 0$.
I first showed that if $\tau^\delta - \tau \in Im \ \sigma$ then $\tau^\delta - \tau = 0 $. To do that I used that every $\sigma(t)$ satisfies $\sigma(t)^\delta = \sigma(t)$ for every $\delta \in \mathbb{S}_n$. If $(\tau^\delta - \tau )^\gamma =\tau^\delta - \tau$ for every $\gamma \in \mathbb{S}_n$ then it's not hard to prove (by taking the powers and using the equations $(\tau^\delta - \tau )^{\delta^j} = \tau^\delta - \tau$ for $ 1 \ \leq j \leq \ order \ of \ (\delta) - 1 ) $ that $\tau^\delta - \tau = 0.$
Now, this only proves that $\{ \tau^\delta - \tau : \tau \in T^k(\mathbb{V}), \delta \in \mathbb{S}_n \} \bigcap Im \ \sigma = 0$. But the problem is that $\Sigma_n$ is the module of all finite sums of elements of the form $\tau^\delta - \tau $ with coefficients in $A$ (you can suppose $A = \mathbb{R}$ a field, since we are really only seeing it for differential manifolds). That is $ \underset{ i=1}{\overset{m}{\sum}}\lambda_i \ .\ (\tau^{\delta_i} - \tau)$.
How can I prove it in general?
The domain splits as the direct sum of the kernel and the image of $\sigma$. Determine the kernel. Then you can note that the map is a surjective homomorphism between vector spaces of the same dimension and hence is injective.