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Let $X$ be a topological space. We introduce a relation $\sim$ on X by declaring $x \sim y$ if for every continuous map $f : X \rightarrow H$ with $H$ a Hausdorff space we have $f (x) = f (y)$. I have shown that

  1. ∼ is an equivalence relation on $X$ and that the quotient space $X/∼$ is Hausdorff.

  2. That there is a functor h: Top → Haus that on objects is given by h(X) = X/∼ By defining the action on arrows like this:

$f: X \rightarrow Y$ is it the unique continuos map $h(f)$ such that $q_Y\circ F=h(f)\circ q_X$ where $q_Y: Y \rightarrow Y/\sim$ and $q_X: X \rightarrow X/\sim$ are the canonical projections

  1. The functor $h$ is left adjoint to the inclusion functor $i:$ Haus → Top. (The space $X/\sim$ is called the maximal Hausdorff quotient of $X$)

I am having trouble proving that the inclusion functor $i:$ Haus → Top does not have a right adjoint. How should I do this?

For this I am supposed to use these properties:

-In Top the monomorphisms are the injective continuous maps whereas the epimorphisms are the surjective continuous maps.

-In Haus a continuous map $f : X \rightarrow Y$ is an epimorphism if and only if f (X) is dense in Y . (In particular, epimorphisms in this case need not be surjective.)

EDIT:

This is an argument of a friend but I am not sure it is ok

Suppose $ f: H_1 \to H_2$ is continuous and H is Hausdorff and $f(H_2)$ is dense in $H$ but $f(H_1)$ is not dense in $H_2$. Then if there were a right adjoint F then the following diagram should commute:

$\text{Hom}_\text{Top}(i(H_1),H)\xrightarrow{\theta_1}\text{Hom}_\text{Haus}(H_2, F(H))$

$\downarrow{-\circ i(f)}\text {.................................. } $ $\downarrow{-\circ f}$

$\text{Hom}_\text{Top}(i(H_1),H)\xrightarrow{\theta_2}\text{Hom}_\text{Haus}(H_2, F(H))$

As $\theta_1\circ f=\theta_2 \circ i(f)$

Choosing $g= id_{H_1}$ will make $\theta_1\circ f$ an epi and $\theta_2 \circ i(f)$ not an epi, since f is not surjective

darkside
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    Well, the hints seem to be leading you towards trying to prove that if a functor has a right adjoint, then it preserves epimorphisms (or using that fact if you've already seen it proved). – Daniel Schepler Oct 23 '23 at 23:21
  • @DanielSchepler I Hven't seen it. I have seen preservation of initial/final objects and products and coproducts only and I don't know how to even start with this. I mean how can the density enter in all this – darkside Oct 23 '23 at 23:32
  • The fact I mentioned isn't very difficult to prove just from the definitions. And then, all you need to do is come up with an example of a morphism which is an epimorphism in Haus but not an epimorphism in Top; this would show that $i$ does not preserve epimorphisms, and then you could apply the contrapositive of the fact. – Daniel Schepler Oct 23 '23 at 23:43
  • @DanielSchepler Do you think the argument I posted in the Edit is correct? If yes, why are θ1∘f an epi and θ2∘i(f) not an epi? – darkside Oct 24 '23 at 05:35
  • @MarianoSuárez-Álvarez Is the argument in the edit wrong? I'd appreaciate some guidance I've been days on this already – darkside Oct 24 '23 at 05:51
  • @PaulFrost Not really, can this argument be fixed with some small modification or is it hopeless? – darkside Oct 24 '23 at 09:13
  • @darkside It can be fixed. See my answer. – Paul Frost Nov 22 '23 at 00:33

1 Answers1

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All you need to know is that a necessary condition for a functor to have a right adjoint is that it preserves epimorphisms. See Question on adjoint functors.

The inclusion functor $Haus \to Top$ does not preserve epimorphisms, thus it does not have a right adjoint.

Paul Frost
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