Right adjoint of functor F: Haus $\rightarrow$ to Top
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 $f (x) = f (y)$. ∼ is an equivalence relation on $X$ and the quotient space $X/∼$ is Hausdorff.
There is a functor $h: Top → Haus$ that on objects is given by $h(X) = X/∼$ and on arrows def by:
$f: X \rightarrow Y$ 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
I have a doubt about my proof that the inclusion functor is right adjoint to h
Proving that the inclusion: $i: $ Haus $\rightarrow$ Top is the right adjoint of h meansproving that
i) $\text{Hom}_{\text{Top}} (X,i(H)) \xrightarrow{\Phi} \text{Hom}_{\text{Haus}}(h(X),H) $ is an isomorphism .
ii) the naturality of the diagram when fixing one of the variables X or H
I think i can define this isomorphism like this:$f \mapsto \bar f$, where $\bar f$ is the quotiented map given by the universal property of the quotient : $f=\bar f \circ q_X$. It is well-defined because $\bar f$ is unique. But I am having trouble proving that it is bijective. I can't invert the map because $q_X$ (which is the canonical projection) is not injective. How do I prove injectivity and surjectivity?
For injectivity I have that if $\Phi (f)=\Phi(g)$ then $\bar f \circ q_X=\bar g\circ q_X$. But I don't know how to conclude that $\bar f = \bar g$
For surjectivity I need that for every $\bar f $ there is an $f$. but the universal property of the quotient says the contrary that for every $f$ there is an $\bar f$
Your definition $$\Phi(f) = \text{ unique map } \bar f : h(x) \to H \text{ such that } f = \bar f \circ q_X$$ is correct.
$\Phi$ is surjective: Given $g : h(X) \to H$, let $G = g \circ q_X$. By definition $\Phi(G)$ is the unique map $\bar G : h(X) \to H$ such that $G = \bar G \circ q_X$. This shows $\Phi(G) = \bar G = g$.
$\Phi$ is injective: Let $f_1, f_2 : X \to H$ such that $\Phi(f_1) = \Phi(f_2)$. This means $f_1 = \Phi(f_1) \circ q_X = \Phi(f_2) \circ q_X = f_2$.