Sums factor as products in internal hom functor

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In case this is helpful, I am working in the category of condensed abelian groups, but I think this can be phrased in a more general categorical way.

Suppose we have a closed symmetric monoidal category $\mathcal{C}$, with internal hom ${\mathscr{H}\kern -.5pt om}$. By the adjunction we have that

$$ {\mathscr{H}\kern -.5pt om} (\lim_{\to} A_i , B) = \lim_{\leftarrow} {\mathscr{H}\kern -.5pt om} (A_i, B)$$

Any direct sum can be written as a filtered colimit (consider finite direct sums). This would mean that

$$ {\mathscr{H}\kern -.5pt om} (\bigoplus A_i, B) = \lim_{\leftarrow} {\mathscr{H}\kern -.5pt om}(A_i, B)$$

One would expect this to be a product but due to my example (more on this below), this does not seem to always be the case. Could it be that the resulting limit just isn't a product?




MY COUNTEREXAMPLE (although I don't think this is strictly necessary to answer the question):

It is not difficult to show that for any set $I$,

$$ \underline{\bigoplus_I \mathbb{Z}} = \bigoplus_I \underline{\mathbb{Z}}$$

where the underline is the standard condensation functor.

Also, for Hausdorff topological groups $A$ and $B$ with $A$ compactly generated, where $\underline{\operatorname{Hom}}(-,-)$ is the internal hom,

$$\underline{\operatorname{Hom}} (\underline{A}, \underline{B}) = \underline{ \operatorname{Hom}( A, B)}$$

where $\operatorname{Hom}(A,B)$ carries the compact-open topology. If sums factored as products,

$$ \underline{\operatorname{Hom}}(\underline{\bigoplus_I \mathbb{Z}}, \underline{\mathbb{Z}}) = \underline{\operatorname{Hom}}(\bigoplus_I \underline{\mathbb{Z}}, \underline{\mathbb{Z}}) = \prod_I \underline{\mathbb{Z}} = \underline{\prod_I \mathbb{Z}} $$

where $\prod_I \mathbb{Z}$ has the discrete topology.

However, we also have

$$\underline{\operatorname{Hom}} (\underline{\bigoplus_I \mathbb{Z}}, \underline{\mathbb{Z}}) = \underline{\operatorname{Hom}(\bigoplus_I \mathbb{Z}, \mathbb{Z})} = \underline{\prod_I \mathbb{Z}}$$

but it is not difficult to see that with the compact-open topology $\prod_I \mathbb{Z}$ does not have the discrete topology.

Of course, there could be some mistake in the above but I can't spot it.

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Just found my mistake. It is not true that

$$\prod_I \underline{\mathbb{Z}} = \underline{\prod_I \mathbb{Z}}$$

The condensation functor commutes with limits of compact Hausdorff spaces, but this is not necessarily the case for discrete groups.