Pip said: “We are trying to show for every x, f(g(x))=f(h(x)), but this shows there is some x1 and x2 such that f(g(x1))=f(h(x2)) and these are different.” Pip still gets confused and is not sure what this shows, so he says, “let’s take your example, g(0)=2 and h(1)=2.... don’t we have to show that g(0) does not equal h(0)?... and then you have to show f(g(0))=f(h(0)) … (he then started to draw Figure 6) and from your example h has to map here (draws h arrow from same point as g arrow in A to different point in B)