The use of universal properties to prove the existence of isomorphism

Question

I just start self learning tensor and I find the universal property is difficult to use.

I think I understand the basic concept of the universal property. The tensor product of $V_1, \cdots, V_m$, $(V,\gamma)$ satisfies $\forall f$, $\exists! T$, s.t. $f=T\gamma$.

\begin{array}{ccc} V_1\times\cdots\times V_m &\rightarrow\gamma & V\\ & \searrow f& \downarrow T\\ && W\\ \end{array}

However, I came across some similar problems which I think they need the same trick on applying universal property, but I cannot figure it out.

Use universal properties to prove there is a canonical isomorphism:

(A) $S_r(V^*)\to S_r(V)^*$, $S_r(V)$ is the r-th symmetric power of V.

(B) $\bigwedge^i(V^*)\to (\bigwedge^iV)^*$

(C) $\bigwedge^2(V\oplus W)\to \bigwedge^2V\oplus(V\oplus W)\oplus\bigwedge^2W$

where $V$, $W$ are both finite dimension vector space.

Could anyone provide me some hint on how to use the universal property to find this isomorphism. I think they must use one common trick to do that, but I think my problem is I cannot find a proper pair like $(V, \gamma)$ to apply the universal property.

Thank you very much!

Answer 1

One important thing you have not written is that though $V_1\times\dots\times V_n$ is a linear space if the $V_i$'s are, the maps $\gamma$ and $f$ from $V_1\times\dots\times V_n$ are not linear maps, but they are mulitlinear: linear in each variable (i.e. if values of the other variables are fixed).

Then the tensor product $V_1\otimes\dots\otimes V_n$ is another vector space, which make multilinear maps like $\gamma$ and $f$ represent as a linear map.

For the exterior power, for example, one can use the alternating multilinear maps.

We have one more important lemma, that we have to use:

If $V$ and $V'$ both satisfy the universal property, then they are isomorphic.

Then, for example, for (B), prove that there is a map $(V^*)^i\to (\Lambda^iV)^*$ which is universal among all alternating multilinear maps $(V^*)^i\to W$.