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Let $X$ and $Y$ be topological spaces. Let $\vert Sing(-)\vert$ be the functor which sends a topological space to the (or "a"? there seem to be more possibilites, for me it's just important, that I get CW complexes at the end) geometric realization of the singular set of the space.

Is the map $C(X,Y)\rightarrow C(\vert Sing(X)\vert,\vert Sing(Y)\vert)$ we get (as $\vert Sing(-)\vert$ is a functor) continuous in the compact-open topology?

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  • $\begingroup$ $|Sing(-)|$ is not even well-defined. If you pick the realizations arbitrarily, how do you extend it to a functor? $\endgroup$
    – Wojowu
    Jan 16, 2019 at 12:04
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    $\begingroup$ @Wojowu: The geometric realization functor is left adjoint to Sing, therefore it is well-defined. $\endgroup$
    – Dmitri Pavlov
    Jan 16, 2019 at 13:57

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This map is not continuous except in some degenerate cases. Take $X = *$ and $Y = [0, 1]$ so that the map reduces to $[0, 1] \to |\mathrm{Sing} [0, 1]|$. Its image is the discrete set of vertices, in particular it is disconnected, so the map is discontinuous.

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