Fix a category $C$. Let $F:C^{op}\rightarrow \operatorname{Set}$ be a functor. For any morphism $T\xrightarrow{u}S$ such that $T\times_ST$ exists, define $$ H^0(S/T,F)\mathrel{\unicode{x2254}} \ker(F(T)\rightrightarrows F(T\times_ST)). $$
Naturally, for another $T'\rightarrow S$ and $S$-morphism $T'\rightarrow T$, this induces a morphism $H^0(T/S,F)\rightarrow H^0(T'/S,F)$.
In Topologies et Faisceaux by Demazure, during the proof of Proposition 1.12, one finds the statement that
By a well-known calculation, the morphism $H^0(T/S,F)\rightarrow H^0(T'/S,F)$ is independent of the morphism $T'\rightarrow T$.
An attempt (Almost a solution)
We have a commutative diagram: $$ \begin{matrix} & & F(S) & & \\ &\swarrow & \downarrow & & \\ H^0(T/S,F) & \rightarrow & F(T) & \rightrightarrows & F(T\times_ST) \\ \downdownarrows & & \downdownarrows & & \downdownarrows \\ H^0(T'/S,F) & \rightarrow & F(T') & \rightrightarrows & F(T'\times_ST') \end{matrix} $$ I think the two compositions $F(S)\rightarrow F(T)\rightrightarrows F(T')$ are naturally equal (we are considering $S$-morphisms), so if we know somehow $F(S)\rightarrow H^0(T/S,F)$ is an epimorphism (or surjective as we are all in the category of sets), then the proof is complete.
Currently I have no idea how to prove this in general; per chance we shall assume that $T\rightarrow S$ has a section? Indeed, if it has a section, then I can show that $F(S)\rightarrow H^0(T/S,F)$ has a section as well.
Since in the context of Proposition 1.12, the morphism $T\rightarrow S$ does have a section, the proof can be said to be complete. But it seems from the text that the result is true in general? (Before this statement one finds the sentence Supposons simplement $T\times_ST$ existe,$\ldots$)
Thanks in advance for any help or reference.
Something is wrong with your link. The text in question is exposé IV of SGA 3.
Consider two $S$-morphisms $f_1,f_2\colon T' \to T$. They induce a canonical morphism $$\alpha = (f_1,f_2)\colon T'\to T\times_S T$$ such that $$f_1 = p_1 \circ \alpha\quad \text{and} \quad f_2 = p_2 \circ \alpha.$$ I denote by $p_1,p_2$ the two canonical arrows $T\times_S T\to T$.
$\mathcal{F}$ induces a morphism $$\mathcal{F} (\alpha)\colon \mathcal{F} (T\times_S T) \to \mathcal{F} (T'),$$ which satisfies $$\mathcal{F} (f_1) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_1) \quad\text{and}\quad \mathcal{F} (f_2) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_2).$$
Now $H^0 (T/S, F)$ is by definition the equalizer of $\mathcal{F} (p_1)$ and $\mathcal{F} (p_2)$, so in particular $\mathcal{F} (f_1)$ and $\mathcal{F} (f_2)$ coincide on $H^0 (T/S, F)$ by the above identities.