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Consider $F: Top \rightarrow Set$ the forgetful functor from the category of topological spaces into the category of sets. I know that its left adjoint is the functor which gives the set a discrete topology. Does this functor have a left adjoint of its own? Intuitively i feel like the answer should be yes, but I have no clue how to prove it.

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  • $\begingroup$ That functor doesn't preserve infinite products, so it can't have a left adjoint. $\endgroup$ Commented Jan 26, 2018 at 2:19
  • $\begingroup$ @DanielSchepler Could you point to a source or some example of a product that isn't preserved? $\endgroup$
    – user525008
    Commented Jan 26, 2018 at 2:24
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    $\begingroup$ For example, the product topology on $\prod_{n=1}^\infty \{ 0, 1 \}$, with the discrete topology on $\{ 0, 1 \}$, is not the discrete topology. $\endgroup$ Commented Jan 26, 2018 at 2:27
  • $\begingroup$ @DanielSchepler And same reasoning could be applied to the right adjoint and coproducts, is that correct? The right adjoint is the functor that gives the set the indiscrete topology, and this doesn't preserve coproducts? $\endgroup$
    – user525008
    Commented Jan 26, 2018 at 2:38
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    $\begingroup$ The discrete space functor does admit a left adjoint, namely the functor of connected components, when you restrict to locally connected spaces. So this lack of a left adjoint is a deficiency, reflecting the presence of pathological spaces, for many purposes. $\endgroup$ Commented Jan 26, 2018 at 16:11

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The discrete topology functor $D$ does not preserve infinite products. For example, the product topology on $\prod_{n=1}^\infty D(\{ 0, 1 \})$ (which gives the categorical product in $\mathbf{Top}$) is not the discrete topology.

It follows that $D$ cannot have a left adjoint.

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