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## When does adding inverses of morphisms preserve commutativity of a diagram?

Here is the essence of a problem I have run in to: I have a finite poset D with a terminal object. If I formally invert all of the morphisms, and add these into my diagram, does the new diagram D' still commute?

I think that the resulting diagram will still commute basically because I have done a lot of examples. Working out a few examples you can see that it basically follows by doing it for the "commutative triangle", and applying this finitely many times. It feels like I should be able to do some kind of messy induction, but I do not really want such a proof cluttering up my work.

Is there a reference I could quote for a result like this? It seems like if it is true it should be a "folk lemma".

Of course, if you have more relaxed criteria for when the result will hold, that would be helpful too.

Also if you know of a conceptual proof which does not fall back on some messy induction, that would be wonderful!

EDIT: An example might help to clarify my question. (How do you draw diagrams on MO?)

 a-->b
^   ^
|   |
c-->d


is my poset. b is the terminal object. Now say someone told you that this was actually a subcategory of a larger category, and in that larger category all of the arrows were invertible. Now consider the larger diagram consisting of the 4 original arrows and their inverses. Is this diagram also commutative? Yes! It is just one or two lines of formal manipulation.

-
 In your diagram above, after inverting you will have two arrows from A to D: one from A to B followed by the formal inverse from B to D, and the formal inverse from A to C followed by the arrow from C to D. To clarify: you're asking that these compositions should coincide? Are you assuming that it's possible to formally invert all your morphisms? – Tom Church Apr 16 2010 at 17:57 Yes, Tom, that is exactly what I mean. I do not seem to be able to express myself very well today. – Steven Gubkin Apr 16 2010 at 17:58

But it should also be easy to prove by induction: for any zigzag of arrows between a and b, the corresponding map in the category with all arrows inverted, when composed with the map from b to the original terminal object, is equal to the map from a to the original terminal object (this is by induction); and so any two maps from a to b in the category with all arrows inverted are equal. In symbols: let me write $t_x$ for the unique morphism in C from $x$ to the terminal object and $[f]$ for the image of $f$ in the category with all arrows inverted. Suppose $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1}$ is a typical map in the category with all arrows inverted with domain $a$ and target $b$. Then the inductive claim is that $[t_b] [f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_a]$, and so $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_b]^{-1} [t_a]$.