**Proof.**
It is clear that (3) implies (2). We will assume (2) and prove (3). The proof is rather formal and we encourage the reader to find their own proof.

Let us first prove that (3) holds when $T_ i$ is in addition assumed separated for all $i$. Choose $i \in I$ and choose a surjective étale morphism $U_ i \to T_ i$ where $U_ i$ is affine. Using Lemma 69.4.2 we see that with $U = U_ i \times _{T_ i} T$ and $U_{i'} = U_ i \times _{T_ i} T_{i'}$ we have $U = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i'}$. Of course $U$ and $U_{i'}$ are affine (see Lemma 69.4.1). Since $T_ i$ is separated, the fibre product $V_ i = U_ i \times _{T_ i} U_ i$ is an affine scheme as well and we obtain affine schemes $V = V_ i \times _{T_ i} T$ and $V_{i'} = V_ i \times _{T_ i} T_{i'}$ with $V = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} V_{i'}$. Observe that $U \to T$ and $U_ i \to T_ i$ are surjective étale and that $V = U \times _ T U$ and $V_{i'} = U_{i'} \times _{T_{i'}} U_{i'}$. Note that $\mathop{\mathrm{Mor}}\nolimits (T, X)$ is the equalizer of the two maps $\mathop{\mathrm{Mor}}\nolimits (U, X) \to \mathop{\mathrm{Mor}}\nolimits (V, X)$; this is true for example because $X$ as a sheaf on $(\mathit{Sch}/S)_{fppf}$ is the coequalizer of the two maps $h_ V \to h_ u$. Similarly $\mathop{\mathrm{Mor}}\nolimits (T_{i'}, X)$ is the equalizer of the two maps $\mathop{\mathrm{Mor}}\nolimits (U_{i'}, X) \to \mathop{\mathrm{Mor}}\nolimits (V_{i'}, X)$. And of course the same thing is true with $X$ replaced with $Y$. Condition (2) says that the diagrams of in (3) are fibre products in the case of $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and $V = \mathop{\mathrm{lim}}\nolimits V_ i$. It follows formally that the same thing is true for $T = \mathop{\mathrm{lim}}\nolimits T_ i$.

In the general case, choose an affine scheme $U$, an $i \in I$, and a surjective étale morphism $U \to T_ i$. Repeating the argument of the previous paragraph we still achieve the proof: the schemes $V_{i'}$, $V$ are no longer affine, but they are still quasi-compact and separated and the result of the preceding paragraph applies.
$\square$

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