@OwenBiesel: Hmm, if we change it to be a three argument function, i.e. take Hom(y,y) as well, we can actually get the right number. We just have to "divide" by Hom(y,y) and if we take the sequence 1,0,0....as 1=X, as the zero length path then reflexivity keeps us from having to divide by zero.

Now when I say divide, I really mean "deconvolute". let P X be sequences, such that \$$X_0 = 1\,\text{and}\,P = \hat{P} \ast X\$$. Then,
\$P\_n= \sum\_{k=0}^n \hat{P}\_{n-k}\*X\_k\$

\$P\_n= P\_n\*{X\_0} + \sum\_{k=1}^{n} \hat{P}\_{n-k}\*{X\_k} \$

\$\hat{P}\_n = P\_n - \sum\_{k=1}^{n} \hat{P}\_{n-k}\*X\_k\$

\$\hat{P}\_n = P\_n - \sum\_{k=0}^{n-1} \hat{P}\_{n-1-k}\*X\_{k+1}\$

\$\hat{P}\_n = P\_n - ((\hat{P}\circ(n \mapsto n-1)) \ast (X \circ (k \mapsto k+1))\_n\$

Call this maping \$$(P,X)\mapsto \hat{P}\$$, \$$"\div"\$$.

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) \$