Let $V,W$ be finite dimensional vector spaces, $V^*$ the dual space of $V$ and $\mathrm{Hom} \ (V,W)$ the vector space of all linear transformations from $V$ to $W$. The universal mapping property states the following:

Let $\phi$ be the bilinear map $(v,w) \rightarrow v\otimes w$ of $V \times W$ into $V \otimes W$. Then whenever $U$ is a vector space and $l: V \times W \rightarrow U$ is a bilinear map, there exists a unique linear map $\tilde{l}: V \otimes W \rightarrow U$ so that $\tilde{l}\circ\phi=l$


Let $F: V^* \times W \rightarrow \mathrm{Hom} \ (V,W)$ be the bilinear map defined by $(f,w)(v)=f(v)\cdot w$. Then by the above property there is a unique linear map $\alpha: V^* \otimes W \rightarrow \mathrm{Hom} \ (V,W)$ such that $\alpha\circ \phi=F$.


How can I show that $\alpha$ is an isomorphism?

  • 1
    $\begingroup$ If you want to show those spaces are isomorphic it suffices verifying the universal property for $(Hom(V, W), F)$ since tensor product is unique up to isomorphism. $\endgroup$ – PtF Apr 2 '16 at 17:32

It is reather easy to see that $F$ is surjective. Choose a basis $w_1,\dots,w_m$ for $w$ and consider a linear map $\Phi:V\to W$. Expanding $\Phi(v)=\sum_i\Phi_i(v)w_i$ defines $\Phi_1,\dots,\Phi_m\in V^*$ and one immediately verifies that $\Phi=F(\sum_i\Phi_i\otimes w_i)$.

For injectivity, I think a bit of additional work has to be done. The main point is to show that the elements $f\otimes w$ span $V^*\otimes W$. To do this using the universal property, you can take the subspace $A\subset V^*\otimes W$ spanned by all such elements and the qoutient projection $\pi:V^*\otimes W\to (V^*\otimes W)/A$. Then $\pi$ vanishes on all elements of the form $f\otimes w$ by construction, so by uniqueness in the universal property, it must be the zero map. But since $\pi$ is surjective, this implies that $(V^*\otimes W)/A=\{0\}$ and hence $A=V^*\otimes W$. Having that at hand, it follows readily that for the Basis above, also the elements of the form $f\otimes w_i$ for $f\in V^*$ and $i=1,\dots,m$ span $V^*\otimes W$. This implies that any element of the tensor product can be written as $\sum_i f_i\otimes w_i$ for some $f_i\in V^*$ and the basis elements $w_i$. But then $F(\sum_i f_i\otimes w_i)=0$ easily implies $f_i=0$ for all $i$ and thus injectivity of $F$.

Alternatively, you can use the same line of argument to show directly that choosing bases of $V$ and $W$, the tensor products of the basis elements form a basis for the tensor product.


Consider bases of $V$ and $W$, $\{v_1,\dots,v_m\}$ and $\{w_1,\dots,w_n\}$. For each $i$ and $j$, consider the linear map $$ f_{ij}\colon V\to W, \qquad f_{ij}(v_k)=\begin{cases} w_j & \text{if $k=i$}\\[4px] 0 & \text{if $k\ne i$} \end{cases} $$ It's easy to prove that $\{f_{ij}:1\le i\le m,1\le j\le n\}$ is a basis for $\operatorname{Hom}(V,W)$: this is basically the method for defining the associated matrix to a linear map between finite dimensional vector spaces, given bases.

Let $\{v_1^*,\dots,v_m^*\}$ be the dual basis on $V^*$ and note that $$ \alpha(v_i^*\otimes w_j)=\phi(v_i^*,w_j) $$ and that, for $k=1,2,\dots,m$, $$ \phi(v_i^*,w_j)(v_k)=v_i^*(v_k)w_j=f_{ij}(v_k) $$ so $\alpha(v_i^*\otimes w_j)=f_{ij}$. In particular $\alpha$ is surjective and the set $\{v_i^*\otimes w_j:1\le i\le m,1\le j\le n\}$ is linearly independent. Since it is also clearly a spanning set for $V^*\otimes W$, it is a basis.

Note. The universal property has been used for defining $\alpha$. In case $V$ is not finite dimensional, the map $\alpha$ is not surjective any more; the set $\{f_{ij}\}$ can be defined from bases, but it's not a basis for $\operatorname{Hom}(V,W)$; it's just linearly independent. So the map $\alpha$ is injective in general, but it is surjective if and only if $V$ is finite dimensional.


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