# Proof that $V^k \mapsto W \cong T^{(k, 0)}V \otimes W$

I am very new to all of this, so please point out if I am misunderstanding something crucial. Anyhow, if $$V^k \mapsto W$$ is the space of $$k$$-linear maps from $$V^k$$ to $$W$$, then I am trying to prove that:

$$V^k \mapsto W \cong T^{(k, 0)}V \otimes W$$

where $$T^{(k,0)} = V^* \otimes \dots \otimes V^*$$. From what I understand, this is equivalent to showing that there exists an isomorphism between these spaces. Starting with $$v_1^* \otimes v_2^* \otimes \dots \otimes v_k^* \otimes w \in T^{(k, 0)}V$$, the first that came to my mind was the map $$\psi: T^{(k, 0)}V \otimes W \mapsto (V^k \mapsto W)$$ given by:

$$\psi(v_1^* \otimes v_2^* \otimes \dots \otimes v_k^* \otimes w)(v_1, v_2, \dots , v_k) = v_1^*(v_1)v_2^*(v_2)\dots v_k^*(v_k)w \in W$$

The map given by $$\psi$$ is $$k$$-linear since $$v_i^*$$ are linear. But since $$\psi$$ has to be an isomorphism to prove the initial statement, doesn't this mean that I have to prove that an inverse mapping exists? Basically, going from $$V^k \mapsto W$$ to $$T^{(k, 0)}V \otimes W$$? I just don't see how I can reverse the operation. How do I even begin to get the functions $$v_i^*$$ from the mapping $$V^k \mapsto W$$? I feel like information is lost when multiplying $$v_i^*(v_i)$$.

Any help is greatly appreciated!

• When you define your map $\psi$ (which is correctly an isomorphism) you need to say that it will be extended by linearity (not all elements of $T^{(k,0)}V$ are of the form $v_1^*\otimes\dots\otimes v_k^*$). Can you see now where the information has gone and why you cannot reverse the operation? (BTW, related math.stackexchange.com/questions/2343836/…) Jun 23, 2021 at 9:55
• @Giulio Interesting, from the source I'm reading they never mention how elements in $T^{(k, 0)}V$ can take any other form than $v_1^* \otimes v_2^* \otimes \dots \otimes v_k^*$. Could you give an example of this? Thanks for linking the question, which is very similar to mine. Do I first need to use the universal property to get from $v_1^* \otimes v_2^* \otimes \dots \otimes v_k^* \otimes w$ to $V \times V \times \dots \times V \times W$?
– Max
Jun 23, 2021 at 13:22
• I was assuming that by $T^{(k,0)}$ you meant exactly $V\otimes\cdots\otimes V\otimes W$. In that case the link answers precisely your question. If not, could you tell (or add it directly in the question) which definition of $T^{(k,0)}$ you are using? About an example of an "indecomposable" tensor take $a,b$ linearly independent and consider $a\otimes b+b\otimes a$ for example. Jun 23, 2021 at 16:11
• @Giulio I now understand why not all elements of $T^{(k,0)}V$ can be written on the form I provided, but as you mentioned, this can be worked around. Thanks for explaining this! I also read the question thoroughly and understand its solution, but one fundamental problem for me is that I can't see how $\psi$ is isomorphic (hence injective) when I cannot reverse the operation. Shouldn't $\psi$ being injective mean that an inverse map exists?
– Max
Jun 23, 2021 at 23:33
• The statement you are trying to prove is not true without some finite-dimensionality assumption. Jun 23, 2021 at 23:43