Help to write a proof (category theory diagram) It is known that $f$, $g$, $h$ are isomorphisms.
It is known that $g\circ f = h^{-1}$.
I need to write down the proof of the following theorem. I am an amateur mathematician and am not an expert in category theory but write a book. The book shall contain this proof. Please help with wording the proof and base results and/or references to other books/articles I can use. (I am going to acknowledge you in my book.)
Theorem The diagram is commutative, every cycle in the diagram is an identity.

Related question:
How do we prove commutativity of a diagram?
 A: You can argue reasonably as the commenters did, but that avoids a perfectly good example on which to hone your more general category theory skills.
You're given that $f, g, h$ are isomorphisms, and that $g\circ f = h^{-1}$.  If you're like me, you desire a somewhat more diagrammatic proof.  Here is one I came up with that also demonstrates the drawing capabilities of an app I'm creating called BananaCats.
Note that $g \circ f = h^{-1}$ has an associated commuting triangle.  And recall the definition of identity morphism on $X$, that it takes any function into or out of $X$ to itself when composed, respectively.
So, the idea is to show that the only function $l$ can be is $\text{id}_X$ based solely on that definition, because that $\text{id}_X$ is the unique such function is a basic property of categories.

Above, the two triangles commute by definition, and our proof goal is to show that the square commutes.  Below, check that this diagram also commutes,

Alternatively, invoke the polygon chord lemma.  

Lemma. Given a commuting polygon and a chord drawn that is the composition of some of its sides, then the two subpolygons also commute!

So in that image, clearly we can draw it backwards: first the outer polygon, and then take the composition of the two sides $g\circ f$ which equal $h^{-1}$ by definition.
That lemma probably doesn't exist in literature, but it's the perfect tool for reasoning visually about these category diagrams.
One final application of the lemma to a subpolygon of the previous diagram give:

Now, we have that $k = k \circ (fgh)$.
You must repeat the process for an arrow entering $C$.  Once you do that, you know that $fgh = \text{id}_C$ is indeed true.  Now argue by symmetry that the rest of the cycles also compose to $\text{id}_X$ on their respective objects.  $\blacksquare$
I like this proof way more than the commenters' proofs, because I got to practice actual category theory in proving the statement.  
