Does the left adjoint to the forgetful functor have another left adjoint?

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.

• That functor doesn't preserve infinite products, so it can't have a left adjoint. – Daniel Schepler Jan 26 '18 at 2:19
• @DanielSchepler Could you point to a source or some example of a product that isn't preserved? – user525008 Jan 26 '18 at 2:24
• For example, the product topology on $\prod_{n=1}^\infty \{ 0, 1 \}$, with the discrete topology on $\{ 0, 1 \}$, is not the discrete topology. – Daniel Schepler Jan 26 '18 at 2:27
• @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? – user525008 Jan 26 '18 at 2:38
• 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. – Kevin Carlson Jan 26 '18 at 16:11

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.