Difficulties with right-adjoint-right-inverse in $\mathsf{Top}$ I'm having difficulties with section $9$ of chapter $\mathrm V$ of CWM.
There is the following proposition:

Proposition 1. If $G:\mathsf C\rightarrow \mathsf D$ is a faithful functor, if $\mathsf D$ has equalizers, and if, for each $X\in \mathsf C$, the functor $(U\downarrow X):(\mathsf C\downarrow X)\rightarrow (\mathsf D \downarrow UX)$ has a right-adjoint-right inverse $L$, then $\mathsf C$ has equalizers.
Proof. To get the equalizer of a parallel pair $f,f^\prime :X\rightarrow Y$, apply $U$, take the equalizer $t:S\rightarrow UX$ of $Uf,Uf^\prime$ in $\mathsf D$ and apply $L$; the universal property of the adjunction shows $Lt:LS\rightarrow X$ is an equalizer in $\mathsf C$.

Okay, so apply $L$ to the equalizer $t$ to get $Lt:LS\rightarrow X$. I don't understand two things:

*

*What is the universal property of the adjunction $(U\downarrow X)\dashv L$?

*Just how does it show $Lt$ is an equalizer of $f,f^\prime$ in $\mathsf C$?

I'm confused about the previous section with the subspace topology as well, so I'll ask about it if needed.
Added: In fact, let me ask another question:


*How can one formulate the universal property of the subspace topology in terms of universal arrows and factorization?

 A: So let $G:C\to D$ be a faithful functor, and assume $D$ has equalizers, and that $(G\downarrow x):(C↓x)\to(D↓Gx)$ has a right-adjoint-right-inverse $L$ (meaning that the counit is always the identity). If $f:f':x\to y$ are parallel arrows, we apply $G$, which gives us distinct arrows $Gf,Gf'$, with an equalizer $t:s\to Gx$. Applying $L$ to $t$ gives us $Lt:Ls\to x$. Note that $G(Lt)=(G↓x)(Lt)=t$.
The universal property of the counit $\varepsilon_g=1_g$ means that, given an arrow $k:(Gh:Gz\to Gx)\to(g:y\to Gx)$, there is a unique arrow $k':(h:z\to x)\to(Lg:Ly\to x)$ such that $(G↓x)(k')=Gk'=k$.
Note that $Gf∘GLt=Gf\circ t = Gf'\circ t = Gf'∘GLt$. By faithfulness of $G$, $Lt$ equalizes $f,f'$.
Now if $h:r\to Ls$ equalizes $f,f'$, then there is a unique arrow $k:Gh\to t$ in the category $(D↓Gx)$ since $t$ is an equalizer, thus terminal object in the category of cones over the parallel arrows $Gf,Gf'$. Next, there is a unique arrow $k':h\to Lt$ such that $Gk'=k$. From uniqueness of $k$, we conclude that $k'$ is the only arrow from $h$ to $Lt$.
