Question: The problem is 

[![enter image description here](https://i.stack.imgur.com/8Mn6d.png)](https://i.stack.imgur.com/8Mn6d.png)

I have been able to prove $left hand \rightarrow composite$, but I haven't be able to prove the reverse direction, so if you could point to me in the right direction via hint, I would really appreciate it.

My attempt:

So the situation is as follows: 

[![enter image description here](https://i.stack.imgur.com/71EmS.png)](https://i.stack.imgur.com/71EmS.png)

Given $f: Q \rightarrow B$ and $g: Q \rightarrow D$, we need a unique $h: Q \rightarrow A$ such that 

-$f = h\_1 h $

-$g = h\_6h$

Now, the right hand square is a pull back, hence there exists a unique $k: Q \rightarrow B$ such that

-$h\_2 k = h\_2 f$

-$h\_7 k = h\_5 g$

And as the composite rectangle is a pull back, there exists a uniqu $i: Q \rightarrow A$ such that 

-$(h\_2h\_1)i = h\_2 f$

-$h\_6 i = g$

Note 

$h\_7(h\_1 i) = (h\_7 h\_1) i = (h\_5 h\_6) i = h\_5 g$

$h\_2(h\_1 i) = h\_2 f$.

So by uniqueness of $k$, $k = h\_1 i $

There also exists unique $i': Q \rightarrow A$ such that 

-$(h\_2 h\_1)i' = h\_2 k$

-$h\_6 i' = f$

By uniqueness, $i' = i$

And I'm not sure how to proceed from here. 

Thanks for your help!

Answer: 
You know there is $N$ such that $\forall n \ge N$ you have $d(x\_n, x) <1/2$ and you know that this implies $x\_n=x$ because the metric is discrete. Thus you proved that not only $x\_N=x$, but that FOR ALL $n \ge N$ it happens that $x\_n=x$.

If the sequence is eventually constant, then it literally means that there are $x$ and $N$ such that $x\_n=x$ for every $n \ge N$. So it trivially converges to that $x$.
Dude, as u said earlier $x\_n$ converging to $x$. Now in discrete metric (or in any metric) $d(x,y)=0$ iff $x=y$ hence now for all $n\geq N$ $d(x\_{n},x)<1/2$ so in this discrete metric $d(x,y)=1$ or $0$ so $d(x\_n,x)=0$ so $x\_n=x$ for all $n\geq N$