Interpolating polynomial discrete log

This is taken from page 16 of Stacking Sigmas

Essentially, let $0<t<\ell$ be integer values smaller than a certain prime modulus $q$. We have a set $\mathcal{X}$ with $|\mathcal{X}|=\ell-t+1$, $[k]:=\{1,2,\ldots,k\}$ and $[k]_0:=[k]\cup0$. For a given set $A\subset\mathbb{Z}_q$, define the polynomials $$f_A(X)=\sum_{i\in A}y_i\cdot L_{(A,i)}(X)$$ where $L_{(A,i)}$ is the Lagrange polynomial over A at the $i$ value, i.e. $L_{(A,i)}(i)=1$ and $L_{(A,i)}(j)=0$ for all $j\in A\setminus\{i\}$.

With all this in mind they make the observation that for all $j\in\mathcal{X}$, $f_\mathcal{X}(j)=f_{[\ell]_0}(j)$, which is correct, but SOMEHOW, this implies that $f_\mathcal{X}=f_{[\ell-t]_0}$ "since both are degree $\ell-t<|\mathcal{X}|$ polynomials".

Then there is one more step and the proof is done, both I don't understand how is that those polynomials are the same, I agree that they have the same degree but I don't see how this implies the equality.

Thanks!

A polynomial of degree $d$ is uniquely determined if it is specified at $d+1$ points. A linear polynomial is uniquely determined if specified at two points. The Lagrange interpolation is just the mechanism for determining the polynomial.

Take $d=\ell -t,$ and note that $\mid {\cal X} \mid =\ell-t+1,$ and that the two polynomials match over the set ${\cal X}.$