I am reading a proof of SAFT, where the following fact is used: If category $D$ is well-powered, and $G: D\rightarrow C$ is a functor, then $(A\downarrow G)$ is well-powered, for any $A$ in $C$. The explanation given is the following: the forgetful functor preserves monos (as it creates and so preserves limits) so the subobjects of $(B,f)$ in $(A\downarrow G)$ are those subobjects $B'\rightarrow B$ in $D$ for which $f: A \rightarrow GB$ factors through $GB'\rightarrow GB$. This is certainly true.
My question is: but isn't it enough just to observe that being a monic in $(A\downarrow G)$ is the same thing as being a monic in $D$, with the additional requirement that certain triangles commute? So, given any object $(B, f)$ in $(A\downarrow G)$ the subobjects of $(B, f)$ will just be the monics $m:B'\rightarrow B$ in $D$ that satisfy the additional property as stated above. i.e, it will be a subset of the poset of subojects of $B$ in $D$. I do not see the need for invoking the forgetful functor here.
edit: My reasoning is flawed. See my comment to Kevin below. I still think this can be done from scratch, but I do not see how yet. Here is what I came up with in the meantime: Consider objects $A$ and $B$ and a morphsim $f$ as shown, in any category $C$, and the square \begin{array}{&&} A & \to & A \\ \downarrow & & \downarrow f \\ A & \stackrel{f}{\to} & B \end{array}
It is easy to show that $f$ is monic $\Leftrightarrow $ the square is a pullback. Since the forgetful functor $U$ preserves limits, and therefore pullbacks we have $f$ monic in $(A\downarrow G)\Leftrightarrow $ square is a pullback in $(A\downarrow G) \Leftrightarrow $ the square is a pullback in $D \Leftrightarrow $ $f$ is a monic in $D$.
This seems a little cleaner than using the full force of $U$, i.e. that it creates small limits. Here, we just use the fact that it preserves pullbacks.
It's true that you can argue directly that if $g:(B',f')\to (B,f)$ is a monomorphism in the comma category it's a monomorphism in $D$, since if $gh=gk$ in $D$ then $gh=gk$ in $(A\downarrow G)$ when we take the view $h:(B'',?)\to (B',f')$ and $k:(B'',?)\to (B',f')$...as of the edit, I don't see this argument going through for monos as it does for epis.