The morphism of rings inducing the diagonal morphism. How to see that the diagonal morphism for affine schemes $\Delta: Spec(R) \rightarrow Spec(R \otimes R)$ is induced by the morphism of rings $R \otimes R \rightarrow R$, $r \otimes r' \rightarrow r \cdot r'$? 
I know this is elementary, but I'm stuck and even don't see this for the induced map on the topological spaces.

To be begin with, let's recall how, if $R$ is a ring of functions on a space $X$, then $R\otimes R$ becomes a ring of functions on $X \times X$.  Namely,
if $f(x)$ and $g(x)$ are two functions on $X$, then $f\otimes g$ gets mapped
to the function $(x,y) \mapsto f(x)g(y)$.
Okay, now lets restrict this function to the diagonal.  We then get the function
$$x \mapsto (x,x) \mapsto f(x)g(x) = (fg)(x).$$
Composing these two, we see that if we map $f\otimes g$ to a function on $X\times X$, and then restrict to the diagonal, we obtain the product function $fg$.  This answers your question.
