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Let $V$ and $W$ be vector spaces over some field $F$. Then $\textrm{Hom}(V, W)$ is the set of homomorphisms between them, which itself forms a vector space.

This definition seems clear enough to me, and I have some intuition for what vector space homomorphisms "are," but I'm always confused when "iterated-Hom" spaces appear; i.e., spaces resembling $\textrm{Hom}(\textrm{Hom}(V, W), U)$. Of course this is "just" the set of homomorphisms from one space to another, but this particular form seems to appear often enough that it must have some nice "meaning."

For instance, the second derivative $D^2 f$ of a real-valued map $f \colon \mathbb{R}^n \to \mathbb{R}^m$ can be thought of as a map from $\mathbb{R}^n$ to $\textrm{Hom}({\mathbb{R}^n, \textrm{Hom}(\mathbb{R}^n, \mathbb{R^m})})$. (The latter space is isomorphic to $\mathbb{R}^{n^2 m}$, but that doesn't seem meaningful, or even useful when the vector spaces aren't finite dimensional.) Higher-order derivatives have analogous interpretations.

How should I intuitively think about such spaces?

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    $\begingroup$ The second section of this text discusses this very point. $\endgroup$ – JuliusL33t Jul 21 '18 at 23:11
  • $\begingroup$ Iterated Hom-spaces are just Hom-spaces. $\endgroup$ – Morgan Rodgers Jul 21 '18 at 23:24
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I think the main thing you have to know is that, for any commutative ring $R$, a,d any $R$-modules $M, N, P$, there are isomorphisms:$\DeclareMathOperator{\Hom}{Hom}$ $$\Hom_R(M\otimes_R N,P)\simeq \Hom_R\bigl(M,\Hom_R(N,P)\bigr)$$ and $$\Hom_R(M\otimes_R N,P)\simeq \mathscr L^2_R (M,N;P)$$ (the set of $R$-bilinear maps from $M\times N$ to $P$). As you know, with your notations, the second differential is a bilinear map from $\mathbf R^n\times\mathbf R^n\to\mathbf R^m$.

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