Denote "the" category of sets and functions by S. The hom set of functions from set X to set Y is denoted by S(X,Y). If C is a cartesian closed category denote by C(x,y) the set of morphisms from x to y in C. In such a C there exists a natural bijection between C(x,y) and C(1,y^x). In a sense, y^x reifies inside C the set C(x,y) in S. Both S(1,C(x,y)) and C(1,y^x) are sets, and in particular, if x=1, then both S(1,C(1,y)) and C(1,y^1) are sets. Anyhow, how does the "external" law of composition C(x,y) x C(y,z) ---> C(x,z) in S of C relate to the "internal" law of composition y^x x z^y ---> z^x in C? In summary, does the internal composition "reify" the external composition?

Denote "the" category of sets and functions by $S$. The hom set of functions from set $X$ to set $Y$ is denoted by $S(X,Y)$. 

If $C$ is a cartesian closed category denote by $C(x,y)$ the set of morphisms from $x$ to $y$ in $C$. In such a $C$ there exists a natural bijection between $C(x,y)$ and $C(1,y^x)$. In a sense, $y^x$ reifies inside $C$ the set $C(x,y)$ in $S$. Both $S(1,C(x,y))$ and $C(1,y^x)$ are sets, and in particular, if $x=1$, then both $S(1,C(1,y))$ and $C(1,y^1)$ are sets. 

Anyhow, how does the "external" law of composition $C(x,y) \times C(y,z) \to C(x,z)$ in $S$ of $C$ relate to the "internal" law of composition $y^x \times z^y \to z^x$ in $C$? In summary, does the internal composition "reify" the external composition?