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This question was already discussed here: Why is Hom(V,W) the same thing as V∗⊗W? But, I want to take a different approach in proving that
$$\phi:V^*\otimes W\to \text{Hom}(V,W)$$ is an isomorphism where $V$ and $W$ are finite dimensional vector spaces. Let's fix basis $\{w_1,\dots,w_m\}$, $\{v_1,\dots,v_m\}$, and $\{\alpha_1,\dots,\alpha_m\}$ for $W$, $V$, and $V^*$ correspondingly where $\{\alpha_i\}$ is a dual basis for $\{v_i\}$. Next, let's define $\phi$ on basis tensors $\alpha_i\otimes w_j\in V^*\otimes W$ via $$\phi(\alpha_i\otimes w_j)=f_{\alpha_i}(w_j)$$ where $f_{\alpha_i}(w_j)(v)=\alpha_i(v)w_j$ and extend it linearly since every element $x$ of $V^*\otimes W$ can be written as a finite sum of the basis elements i.e. if $x\in V^*\otimes W$, then $x=\sum_{i,j}a_{ij}\alpha_i\otimes w_j$ and $$\phi(x)=\sum_{i,j}a_{ij}\phi(\alpha_i\otimes w_j)=\sum_{i,j}a_{ij}f_{\alpha_i}(w_j).$$

Next, I claim that it is enough to show that $\ker(\phi)=0$ since $V^*\otimes W$ and $Hom(V,W)$ have the same dimension and $\phi$ is a linear map (rank-nullity Theorem).

So, let's show that $\ker(\phi)=0$. We can do that by fixing the basis for $V$, $W$, and $V^*$ and showing that if $\phi(x)=0$ then $x=0$ (Details are skipped due to computations).

Does it make sense?

A follow-up question: Do we have an isomorphism if we take either of these spaces to be infinite dimensional? What would we do in that case?

eightc
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    Yes, it makes sense. No, we don't have the isomorphism in infinite-dimensional case, the reason is that not every linear operator between infinite-dimensional vector spaces has finite rank. – Mykola Pochekai Sep 08 '20 at 23:14
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    No, it does not make sense: your reasoning has an important gap. You can not just make up a formula on all elementary tensors and extend “by linearity” to all tensors because elementary tensors span but have lots of linear relations. You need to observe that your formula on elementary tensors is bilinear in $\alpha$ and $w$ (think of defining a bilinear map on $V^* \times W$ first and then factoring it through the tensor product). For example, you can’t define $f : V^* \otimes W \to V^* \oplus W$ by $f(\alpha \otimes w) = (\alpha,w)$ and then “extend by linearity”, right? – KCd Sep 08 '20 at 23:21
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    Only $V$ needs to be finite-dimensional. – Qiaochu Yuan Sep 08 '20 at 23:26
  • Thank you for your comment! That was helpful. I think I can fix that problem by fixing the basis and defining my map on the basis elements. Can you check that please? – eightc Sep 09 '20 at 00:03
  • Also, here https://math.stackexchange.com/questions/2742627/is-homv-w-simeq-v-otimes-w-naturally?rq=1 they defined the map on a tensor $\alpha\otimes w$. I am a little bit confused right now. – eightc Sep 09 '20 at 00:06
  • @KCd the OP does say "let's define $\phi$ on basis tensors $\alpha_i\otimes w_j\in V^*\otimes W$ and [...] extend" - the key word being "basis", of course. – peter a g Sep 09 '20 at 01:11
  • @peterag Ah, I missed that. I'd say giving a formula on "basis tensors" first is a bad way of constructing a linear map out of a tensor product space, since it is important that its value on all elementary tensors is given by the indicated expression in the OP's post. This could be shown as a consequence of the formula on "basis tensors" first, but that's clunky compared to the more elegant technique of constructing a linear map with values specified on all elementary tensors initially and then followed by extension to all tensors by the univ. mapping property (unclear of OP knows that). – KCd Sep 09 '20 at 01:46
  • @KCd right! I just wanted the OP not to be confused, especially as your comment got 4 "likes"... – peter a g Sep 09 '20 at 01:54
  • Thank you everyone for your feedback! – eightc Sep 09 '20 at 02:06

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