Why does the inner square commute if all outer squares commute? In proving a change of basis theorem in linear algebra, our professor draw this diagram and simply stated that because all the  outer squares in this diagram commute, the inner square (green) must also commute (I didn't write the exact mappings, because I think this question is more about diagram chasing and that it isn't really relevant).
I can't, however, figure out why this is true. This class is also my first time experiencing commutative diagrams, so please explain it with as many details as possible.


Edit: it turned out the questions wasn't fine, since "in general there is no implication either way", but "if the diagonal arrows are isomorphisms then the inner square commutes if and only if the (big) outer square commutes".
In my case, the diagonal arrows actually are isomorphisms, which is why I am posting the extended diagram.

To clear some notation:

*

*$A: U \to V$ is a linear map between vector spaces $U$ and $V$. First bases for $U, V$ are $B, C$. Another possible bases for $U, V$ are $B', C'$.

*$\phi$'s are isomorphisms

*$P_{XY}$ is a change-of-basis matrix from $Y$ to $X$
I would still appreciate if someone would help me understand why the inner square commutes, because I am lost.
 A: 
because all the outer squares in this diagram commute, the inner square (green) must also commute

That is not true in general. You can make a counterexample in the category of vector spaces by letting the green square be your favorite non-commuting square and then declare that the four outer objects in the black square are all the trivial space $\{0\}$. Then the outer squares will commute automatically, because linear transformations always take $0$ to $0$.
So you need to know more about the maps before you can conclude the green square commutes.

After question was updated: We now know that the blue arrows are isomorphisms -- then it's quite different.
A useful principle is: if you have a commuting square (or other diagram) and replace an isomorphism with its inverse, the resulting diagram still commutes.
In your case, you can flip just $\Phi_B$ in your diagram, and you should now be able to show step by step that the green square commutes.
(In algebraic notation, what's going on is just that, for example, $A_{CB}\circ \Phi_B = \Phi_C\circ A$ implies $A_{CB} = \Phi_C \circ A \circ \Phi_B^{-1}$ when we compose with $\Phi_B^{-1}$ on the right.)
A: Allow me to change notation.
Essentially, you are in this situation:
\begin{align}
f' \circ a & = b \circ f \\
g' \circ b & = d \circ g \\
h' \circ a & = c \circ h \\
k' \circ c & = d \circ k
\end{align}
You want to know if $g \circ f = k \circ h$ implies $g' \circ f' = k' \circ h'$.
In general, there is no implication.
Certainly, if $g \circ f = k \circ h$ then:
\begin{align}
g' \circ f' \circ a 
& = g' \circ b \circ f \\
& = d \circ g \circ f \\
& = d \circ k \circ h \\
& = k' \circ c \circ h \\
& = k' \circ h' \circ a
\end{align}
So if we could cancel $a$ on the right then we would be done.
In other words, if you additionally assume that $a$ is an epimorphism then $g \circ f = k \circ h$ implies $g' \circ f' = k' \circ h'$.
By duality, if you assume that $d$ is a monomorphism then $g' \circ f' = k' \circ h'$ implies $g \circ f = k \circ h$.
In particular, if $a$ and $d$ are isomorphisms then $g \circ f = h \circ k$ if and only if $g' \circ f' = h' \circ k'$.
