TITLE: When does adding inverses of morphisms preserve commutativity of a diagram?
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: There is an easy conceptual proof using the fact that the category obtained by formally inverting all the arrows in a category C is equivalent to the fundamental groupoid of the nerve NC of C, and that the nerve of a category with a final object is contractible.  Without the assumption of a final object your assertion is false in general, e.g., reverse the arrows from c in your example.
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 $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_b]^{-1} [t_a]$.