Map that sends one map to another Question asks to prove that $(g  \circ h)_* = g_* \circ h_*$ and $(g \circ h)* = h^* \circ  g^*$, where $f_*=f \circ$ and $f^* = \circ f$. So it is basically some operation that sends one map to other. I did not understand the meaning $(g_* \circ h_*)$. I can see that $(g \circ h)_*(X \rightarrow Y))$ Just takes a $g(h(X \rightarrow Y))$. But how do we define $g_* \circ h_*$ for $X \rightarrow Y$ as an element?
 A: I think you need to reconsider your notation!
You have given $2$ maps:
$$X \xrightarrow{h} Y  \xrightarrow{g} Z$$ 
Now lets consider $Set(\_ , M)$, thwe set of maps from some set to a given set $M$ (I call it Set(/_ , /_) because it is the morphism set in the category of sets).
Now every map $f: Z \to M$ induces a map $f\circ g : Y \to M$. So we get a map $$h^*:Set(Z,M) \to Set(Y,M)\\ f \mapsto f \circ h$$ And analogously for $Y,Z$ and $h$ (observe that here the arrows "flip"). This map is referred to as $g^*$ respectivel $h^*$ so you need to prove, that $$g^* \circ h^* (f:Z \to M )=g^*(h^*(f: Z \to Y))\stackrel{!}{=}(g\circ h) ^*$$ which just boils down to unfolding the definition.
Now gfor the $g^*$ we just do the same thing dually:
I.e. we now consider $Set(M,\_)$ here we now get the maps $h^*$ $g^*$ in exactly the same way (only that now the arrows don't get "reversed") since we postcompose. I.e. 
$$h_*:Set(M,Y) \to Set(M,Z)\\ f \mapsto h \circ f$$
I hope that helps your understanding, writing down more details might be even more confusing for you, and with these thingys it really helps to once wrap your head around yourself! 
