TITLE: In what sense is the S-combinator "substitution"?
QUESTION [6 upvotes]: According to the Wikipedia page on SKI-combinator calculus, I is the identity function, K is the constant function, and S is "substitution".  I understand the first two, but I don't see what S has to do with substitution and wikipedia offers no explanation or clarification.

REPLY [5 votes]: I think the name $S$ arises because of the role that the $S$ combinator plays in the conversion from lambda-terms.  Suppose you have a lambda-term $\lambda a.(P Q)$ and you want to find an equivalent $SKI$-combinator. (The $S$ combinator is not involved except when the term has this form.)  This term  takes its argument   $T$ and substitutes it into $P$ and $Q$ in place of the free variable $a$, then applies the result $P[a:=T]$ to the result $Q[a:=T]$.
When we convert to combinatory logic, there are no parameter variables.  So in place of the $\lambda$-abstraction, we need a combinator which performs an analogous substitution operation on its arguments. This combinator, which we  call $S$, will take $P$ and $Q$ and an argument $T$ and perform the combinatorial analogue of substitution before applying $P$ to $Q$.  That is, we want $$S P Q T = (P T) (Q T).$$
The other two $SKI$-combinators, $K$ and $I$, do not perform this substitution.  $I$ is used in the trivial case when one is converting the term $\lambda a.a$, and $K$ is used in the case where there is no substitution, when the bound variable $a$ does not appear in the body of the abstraction and so no substitution of the argument for the bound variable is necessary.
(I have been trying to find a citation for this explanation, which I think I did not make up myself, but not had any success so far.  If I do find one I will add it.)
[Added 2021: It appears that the letter “S” was originally chosen not for its association with substitution, but because it was the first letter of schmelzen, “to fuse”, and that Schönfinkel understood the $S$ combinator not as a substitution operator but as a fusion operator. See my remarks in the other answer.]