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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!

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    $\begingroup$ 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/…) $\endgroup$
    – Giulio R
    Jun 23, 2021 at 9:55
  • $\begingroup$ @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 $V \times V \times \dots \times V \times W$? $\endgroup$
    – Max
    Jun 23, 2021 at 13:22
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    $\begingroup$ 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. $\endgroup$
    – Giulio R
    Jun 23, 2021 at 16:11
  • $\begingroup$ @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? $\endgroup$
    – Max
    Jun 23, 2021 at 23:33
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    $\begingroup$ The statement you are trying to prove is not true without some finite-dimensionality assumption. $\endgroup$
    – Eric Wofsey
    Jun 23, 2021 at 23:43

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