I am reading the proof of a necessary and sufficient condition for a presheaf over a base of a topology to be a sheaf, from Élements de Géometrie Algébrique I, chapter 0, point 3.2.2. There is a line where I'm lost and I can't understand where the things the author says come from.
Context
Let $X$ be a topological space and $B$ a base of its topology, considered as a category with inclusion maps as morphisms. A presheaf on $B$ is just a contravariant functor $\mathcal{F}:B\to \mathcal{C}$, where $\mathcal{C}$ is any category that admits projective limits. Denote by $\rho^U_V:\mathcal{F}(U)\to\mathcal{F}(V)$ the restriction morphisms. From this we can define a presheaf $\mathcal{F}'$ on $X$ by setting $\mathcal{F}'(U)=\varprojlim\mathcal{F}(V)$, where $V\subseteq U$ and $V\in B$.
In the proof of 0-3.2.2, the author bassicaly shows that this definition doesn't depend on $B$ and therefore it fulfills the condition needed to be a sheaf with the following hypothesis:
For every covering $(U_\alpha)$ of $U\in B$ by sets $U_\alpha\in B$ contained in $U$, and for every object $T\in\mathcal{C}$, the map that sends every $f\in\mathrm{Hom}(T,\mathcal{F}(U))$ to the family $(\rho^U_{U_\alpha}\circ f)\in\prod\mathrm{Hom}(T,\mathcal{F}(U_\alpha))$ is a bijection onto the the family $(f_\alpha)$ such that $\rho^{U_\alpha}_V\circ f_\alpha=\rho^{U_\beta}_V\circ f_\beta$ for every pair of indices $(\alpha,\beta)$ and every $V\in B$ with $V\subseteq U_\alpha\cap U_\beta$.
In order to do that, he chooses a base $B'\subseteq B$, and defines $\mathcal{F}''$ in the same way as $\mathcal{F}'$ but taking projective limit over the elements of $B'$.
Question
For every open set $U$ there is a morphism $\mathcal{F}'(U)\to\mathcal{F}''(U)$ which is the projective limit of $\mathcal{F}'(U)\to\mathcal{F}(V)$ for $V\in B'$. Now, if $U\in B$, we want to show that $\mathcal{F}'(U)\to\mathcal{F}''(U)$ is an isomorphism, and the following is stated
il est immédiat de voir que les composés des morphismes $\mathcal{F}(U)\to\mathcal{F}''(U)$ et $\mathcal{F}''(U)\to\mathcal{F}(U)$ ainsi définis sont les identités.
In english:
it is immediate to see that the compositions of the morphisms $\mathcal{F}(U)\to\mathcal{F}''(U)$ and $\mathcal{F}''(U)\to\mathcal{F}(U)$ defined like this are identities.
What I don't see is how is $\mathcal{F}(U)\to\mathcal{F}''(U)$ defined. The other direction is straight forward from the universal property of the projective limit, but there is no clue in the previous text of how $\mathcal{F}(U)\to\mathcal{F}''(U)$ is defined (or at least I can't find it).
Alternative
Alternatively, I tried to prove the theorem using the morphisms $\mathcal{F}'(U)\to\mathcal{F}''(U)$ and back, given by the filtering property of any base. But I don't know how to show that these compositions are identities. I would consider my question answered if this alternative is answered.
I found a way to prove the statement using the alternative idea.
Let's call $f:\mathcal{F}'(U)\to\mathcal{F}''(U)$ and $g:\mathcal{F}''(U)\to \mathcal{F}'(U)$. Then, by commutativity we can decompose $\rho_{UV}:\mathcal{F}'(U)\to\mathcal{F}(V)$ as $\mathcal{F}'(U)\to \mathcal{F}''(U)\to\mathcal{F}'(U)\to \mathcal{F}(V)$, which is $fg\circ\rho_{UV}$. Therefore we have $\rho_{UV}\circ id_V=fg\circ\rho_{UV}$ for every $V$. Using the hypothesis, $fg$ must be the identity, and therefore we have the isomorphism.