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gave Schwartz kernel theorem some minimal content
I am wondering now if we may lift the Schwartz kernel theorem to morphisms in $\mathbf{H} =$ smooth sets or in $\mathbf{H}$ the Cahier topos, under the identification discussed at distributions are the smooth linear functionals.
So conjecturally $\mathbb{R}$-linear morphisms in $\mathbf{H}$ of the form
$\left[ \mathbb{R}^n, \mathbb{R} \right] \otimes_{\mathbb{R}} \left[ \mathbb{R}^n, \mathbb{R} \right] \longrightarrow \mathbb{R}$should be in linear bijection with $\mathbb{R}$-linear morphisms in $\mathbf{H}$ of the form
$\mathbb{R}^n \times \mathbb{R}^n \longrightarrow \mathbb{R}$The would-be proof would first observe that by Cartesian closure the former are equivalent to $\mathbb{R}$-linear morphisms of the form
$[\mathbb{R}^n, \mathbb{R}] \longrightarrow \left[ [\mathbb{R}^n, \mathbb{R}], \mathbb{R} \right]_{\mathbb{R}} \,.$Using on the right that distributions are the smooth linear functionals this is smooth functions from the space of smooth functions to that of compactly supported distributions.
Now IF that identification would also apply here (i.e. where the codomain is not $\mathbb{R}$, but a space of distributions), we would conclude that this is the same as the continuous linear functions
$\mathcal{E}(\mathbb{R}^n) \longrightarrow \mathcal{E}'(\mathbb{R}^n)$The Schwartz kernel theorem would then identify this with $\mathcal{E}'(\mathbb{R}^n \times \mathbb{R}^n)$ which, once more, is the same as $\mathbb{R}$-linear morphisms in $\mathbf{H}$ of the form.
$\left[\mathbb{R}^n \times \mathbb{R}^n , \mathbb{R} \right] \longrightarrow \mathbb{R}^n \,.$Hence the question is:
Does the proof of
$\mathbf{H}\left( [\mathbb{R}^n, \mathbb{R}], \mathbb{R} \right)_{\mathbb{R}} \;\simeq\; \mathcal{E}'(\mathbb{R}^n)$generalize to yield
$\mathbf{H}\left( [\mathbb{R}^n, \mathbb{R}],\,\, [[\mathbb{R}^n, \mathbb{R}],\mathbb{R}]_{\mathbb{R}} \right)_{\mathbb{R}} \;\simeq\; \mathcal{E}'(\mathbb{R}^n, \mathcal{E}'(\mathbb{R}^n))$??
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