# Is the map between mapping spaces, induced by the functor $\vert Sing(-)\vert$ continuous?

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?

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

## 1 Answer

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.