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.

>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.