How does an Object at event horizon stop to the observer? I have read about an object, that once at the even horizon the object would be seen as stopped to the observer.
So my question is in regards to the light reflection off of the object. If the object has stopped and my understanding is the light itself is unable to escape, how does the light reflect off of the object for eternity in order for the observer's eyes to receive.
If it is due to the light on the facing side is still able to escape, then my thought is the object surface facing the observer would reduce in area unil only the last surface able to reflect light would be available.  Thus the item would result in a dot of light reflection.
I hope this question is clear and is sensible.
 A: At a correct moment enough light to last for an eternity was send after the object. A one second pulse for example, and infinite brightness of course.
The front part of the pulse arrives back after 10 years, the middle part arrives back after 1000 years, the rear end arrives back after eternity.
The middle part will be seen being reflected from an object that was slightly above the horizon.
And let's formulate the thing about light not escaping from the horizon like this: It takes an eternity for the light to escape the horizon.
A: Instead of dealing with light being reflected off the falling body, I'd like to give the falling body a flashlight, so we won't deal with ingoing and outgoing light rays, only outgoing.
I'll give you both intuitition and math on this question.
Firstly, you have a misconception about black holes. A far observer never sees the falling body touch the event horizon. As the falling body falls, its time dilation increases. It takes an infinite amount of time for the object to reach the event horizon according to the far observer. And as long as the flashlight did not reach the event horizon, you can see its light.
If you want a mathematical part, the Schwarzschild metric is given by (neglecting angles):
$$ds^2 = (1-\frac{r_s}{r})dt^2 - (1-\frac{r_s}{r})^{-1}dr^2$$
Light rays move such that their proper distance $ds=0$.
substituting, you find out that:
$$dt=\pm \frac{rdr}{r-r_s}$$
Taking the positive solution for the outgoing light ray:
$$dt= \frac{rdr}{r-r_s} \implies t=r+r_s\ln|r-r_s|+c$$
where I got the second result by integrating and $c$ is an integration constant.
Graphically, it looks like this:
Where the vertical axis is time, the horizontal axis is the distance, the black region is inside of the Event Horizon, and the red lines are the light rays.
You can see that near the black hole, light rays are not moving quickly away from the event horizon, but as long as they did not start out from the event horizon, they leave the black hole.
So in other words, you still see the falling body forever, because it taken infinite time for him to reach the event horizon.
A: The fundamental reason for this - i.e. independent of coordinates and observation methods - is the "one-way" nature of the horizon, in that information cannot flow from the interior region to the exterior.
Consider the counterfactual: suppose that you could see the object cross the horizon after a finite time. Then by combining your information from before and the new information from after, you could make a logical inference that within the horizon volume there is now an object. That means you now have a piece of information about what is going on inside part of the Universe you are not "supposed" to be able to have information about the goings-on in.
Hence, something must happen so as to make your reasoning fail. And the way the real world does this is by dragging out the light signals coming from that object with more and more "latency", becoming an infinite amount of such latency for a signal emitted right at the horizon. You thus never get to establish the truth of one of your premises, and your argument fails to go through. And so you gain no valid information about what is inside the horizon. This is not the only way the information could be blocked, but it is the way that our world does it.
