MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?)

 ^   ^
 |   |

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.

share|cite|improve this question
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 '10 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 '10 at 17:58
up vote 5 down vote accepted

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 $[f_1]^{\pm 1} \cdots [f_n]^{\pm 1} = [t_b]^{-1} [t_a]$.

share|cite|improve this answer
Awesome. I considered the counterexample to the general version of the question, and this is why I added the existence of a terminal object to my suppositions. So your conceptual proof fails for this example because you end up with a circle which is not contractible! Pretty neat! I accepted you answer for the conceptual treatment, but I still do not quite see the induction. What are you inducting over exactly? I must be really dense today... – Steven Gubkin Apr 16 '10 at 18:13
I explained more clearly what the inductive claim is. – Reid Barton Apr 16 '10 at 18:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.