Euler Characteristic is not Dependent upon Triangulation

Theorem

Let $S$ be a surface.

Let $T_1$ and $T_2$ be triangulations of $S$.

Let:

$\map {\chi_1} S$ be the Euler characteristic of $S$ as calculated using $T_1$

$\map {\chi_2} S$ be the Euler characteristic of $S$ as calculated using $T_2$.

Then:

$\map {\chi_1} S = \map {\chi_2} S$

Proof

Recall:

$\ds \map {\chi_1} S$

$=$

$\ds \map {\chi_2} S$

$\ds \leadstoandfrom \ \ $

$\ds \map v {T_1} - \map e {T_1} + \map f {T_1}$

$=$

$\ds \map v {T_2} - \map e {T_2} + \map f {T_2}$

Definition of Euler Characteristic of Surface

where:

$\map v {T_1}$ denotes the order of $T_1$ (that is, the number of its vertices)

$\map e {T_1}$ denotes the size of $T_1$ (that is, the number of its edges)

$\map f {T_1}$ denotes the number of faces of $T_1$.

This theorem requires a proof.
In particular: Probably do it by induction on number of faces, once we have a rigorous definition of triangulation, and have the basics of the theory down.
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Sources

2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Euler characteristic