Does the forgetful functor $\mathbf{Comp} \rightarrow \mathbf{Top}$ have a left-adjoint?

The forgetful functor $\mathbf{CompHaus} \rightarrow \mathbf{Top}$ from compact Hausdorff spaces to topological spaces famously has a left-adjoint, the Stone-Cech compactification.

Question. Does the forgetful functor $$\mathbf{Comp} \rightarrow \mathbf{Top}$$ from compact spaces to topological spaces have a left-adjoint?

This should be well-known, but I haven't been able to find any relevant information.

• Remark. It seems unlikely, if such a functor exists, that it will preserve Hausdorffness. I'm completely okay with this. – goblin Jul 10 '17 at 18:06
• CompHaus to Top indeed has a left adjoint, so why is this not a left adjoint for Comp to Top? What goes wrong? – Henno Brandsma Jul 10 '17 at 19:34
• @HennoBrandsma By full-faithfulness of the inclusion functor $i$, the counit $L i X \to X$ would have to be an isomorphism. But if $L$ lands us in compact Hausdorff spaces, then each such $X$ would be compact Hausdorff. – Todd Trimble Jul 10 '17 at 19:41

No, it does not. If it did, then $\mathbf{Comp}$ would be a reflective subcategory of the total category $\mathbf{Top}$, and hence would be total itself. Now, total categories admit all small limits, and there is the rub: $\mathbf{Comp}$ does not have all small limits. In fact, $\mathbf{Comp}$ does not have equalizers (of course it admits small products, by the Tychonoff theorem).
Here is a simple example to show why not. Consider Sierpinski space $\mathbf{2} = \{0, 1\}$ where the point $1$ is open. This is of course compact. Take two continuous maps $f, g: [0, 1] \rightrightarrows \mathbf{2}$ where $f^{-1}(1) = [0, 1/2)$ and $g^{-1}(1) = [0, 1]$. Suppose there is an equalizer $i: E \to [0, 1]$ of $f, g$ in $\mathbf{Comp}$. Since $\hom(1, -): \mathbf{Comp} \to \mathbf{Set}$ would preserve this equalizer, we see the underlying set of $E$ would have to be $[0, 1/2)$. In order for $i: E \to [0, 1]$ to be continuous, the topology on $E = [0, 1/2)$ would have to be equal to or finer than the subspace topology on $[0, 1/2)$. But since the subspace topology on $[0, 1/2)$ is not compact, no finer topology on $[0, 1/2)$ can be either; contradiction.