Christopher Upshaw wrote:
>lets name this not-quite-an-enriched-category \$$\mathscr{P}\$$ for "Paths", and use that as well for the function from pairs of objects to sequences. Suppressing what graph it is made from for now
\$\text{At B} ,\mathscr{P}(A,B) \otimes \mathscr{P}(B,C) = (\mathscr{P}(A,B)\div\mathscr{P}(B,B))\ast\mathscr{P}(B,C) \$
And then
\$\mathscr{P}(A,B) \otimes \mathscr{P}(B,C) \le \mathscr{P}(A,C) \$

Nice! I see what you mean by it being not-quite an enriched category, since there's a different tensor product operation for composition at each object.

But what if we did something slightly different: what if the enrichment sends a pair of objects \$$A,B\$$ not to the sequence of path lengths \$$\mathscr{P}(A,B)\$$ but to the "reduced" sequence \$$\mathscr{R}(A,B) = \mathscr{P}(A,B) \div \mathscr{P}(B,B)\$$? I think this sequence counts something like "Paths from \$$A\$$ to \$$B\$$ of length \$$n\$$ that don't meet \$$B\$$ before the end of the path."

Regardless, then we'd have (assuming that if \$$N\leq M\$$ then \$$N\div P \leq M\div P\$$)

\$\mathscr{R}(A,B) \ast \mathscr{R}(B,C) = \mathscr{P}(A,B) \div \mathscr{P}(B,B) \ast \mathscr{P}(B,C) \div \mathscr{P}(C,C)\$
\$= (\mathscr{P}(A,B) \ast \mathscr{P}(B,C) \div \mathscr{P}(B,B)) \div \mathscr{P}(C,C) \$
\$\leq \mathscr{P}(A,C) \div \mathscr{P}(C,C)= \mathscr{R}(A,C).\$

Something tells me this can't work, though. If \$$\mathscr{R}(A,B)\$$ has the interpretation I mentioned, then if there exist paths \$$A\to B\$$ and \$$B\to A\$$ we must have \$$\mathscr{R}(A,B)\ast\mathscr{R}(B,A) \leq \mathscr{R}(A,A) = (1,0,0,\dots)\$$, which doesn't make sense.