= Solution
The <Papperitz symbol>
$$
P\left\{
\begin{array}{ccccc}
0&1&\infty&&\\
0&0&a&z&\\
1-c&c-a-b&b&&
\end{array}\right\}
$$
specifies a second-order <Fuchsian differential equation>. The first row lists its three distinct regular singular points $0,1,\infty$; $z$ is the independent variable. The two entries below each singular point are its <characteristic exponent at a regular singular point>[characteristic exponents]. Thus local solutions have leading behaviors
$$
1,\ z^{1-c}\quad(z\to0),
$$
$$
1,\ (1-z)^{c-a-b}\quad(z\to1),
$$
and, with the usual convention at infinity,
$$
z^{-a},\ z^{-b}\quad(z\to\infty).
$$
When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the <Fuchs relation>
$$
0+(1-c)+0+(c-a-b)+a+b=1.
$$
For a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the <Gauss hypergeometric equation>. Its distinguished solution $F(a,b;c;z)$ is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes $c\notin\{0,-1,-2,\ldots\}$, with exceptional parameter values handled separately or by continuation.
Now put
$$
u=\frac{z}{z-1},
\qquad
Y(z)=F(a,c-b;c;u).
$$
The hypergeometric equation in $u$ has exponent pairs
$$
(0,1-c)\text{ at }u=0,\qquad
(0,b-a)\text{ at }u=1,\qquad
(a,c-b)\text{ at }u=\infty.
$$
The <Möbius transformation of a Papperitz symbol> sends
$$
u=0,1,\infty
\quad\longleftarrow\quad
z=0,\infty,1,
$$
so $Y$ has symbol
$$
P\left\{
\begin{array}{ccccc}
0&1&\infty&&\\
0&a&0&z&\\
1-c&c-b&b-a&&
\end{array}\right\}.
$$
On the other hand, $F(a,b;c;z)$ has exponent pairs
$$
(0,1-c),\qquad(0,c-a-b),\qquad(a,b)
$$
at $0,1,\infty$. Multiplication by $(1-z)^a$ applies the <dependent-variable rescaling of a Papperitz symbol>: it adds $a$ to both exponents at $z=1$ and subtracts $a$ from both at infinity. Hence
$$
(1-z)^aF(a,b;c;z)
$$
has exactly the same three exponent pairs as $Y$.
Both functions are analytic near $z=0$ and equal one at $z=0$, so uniqueness of the normalized exponent-zero hypergeometric solution gives the <Pfaff transformation>
$$
\boxed{
F\left(a,c-b;c;\frac{z}{z-1}\right)
=(1-z)^aF(a,b;c;z).}
$$
The identity first holds near zero with the branch of $(1-z)^a$ equal to one there, and then extends by <analytic continuation> on any domain where compatible branches are chosen.
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