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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 $mathcal{F}:Bto mathcal{C}$, where $mathcal{C}$ is any category that admits projective limits. Denote by $rho^U_V:mathcal{F}(U)tomathcal{F}(V)$ the restriction morphisms. From this we can define a presheaf $mathcal{F}'$ on $X$ by setting $mathcal{F}'(U)=varprojlimmathcal{F}(V)$, where $Vsubseteq U$ and $Vin B$.

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 $Uin B$ by sets $U_alphain B$ contained in $U$, and for every object $Tinmathcal{C}$, the map that sends every $finmathrm{Hom}(T,mathcal{F}(U))$ to the family $(rho^U_{U_alpha}circ f)inprodmathrm{Hom}(T,mathcal{F}(U_alpha))$ is a bijection onto the the family $(f_alpha)$ such that $rho^{U_alpha}_Vcirc f_alpha=rho^{U_beta}_Vcirc f_beta$ for every pair of indices $(alpha,beta)$ and every $Vin B$ with $Vsubseteq U_alphacap U_beta$.

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 $mathcal{F}'(U)tomathcal{F}''(U)$ which is the projective limit of $mathcal{F}'(U)tomathcal{F}(V)$ for $Vin B'$. Now, if $Uin B$, we want to show that $mathcal{F}'(U)tomathcal{F}''(U)$ is an isomorphism, and the following is stated

il est immédiat de voir que les composés des morphismes $mathcal{F}(U)tomathcal{F}''(U)$ et $mathcal{F}''(U)tomathcal{F}(U)$ ainsi définis sont les identités.

In english:

it is immediate to see that the compositions of the morphisms $mathcal{F}(U)tomathcal{F}''(U)$ and $mathcal{F}''(U)tomathcal{F}(U)$ defined like this are identities.

What I don't see is how is $mathcal{F}(U)tomathcal{F}''(U)$ 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 $mathcal{F}(U)tomathcal{F}''(U)$ is defined (or at least I can't find it).

Alternative

Alternatively, I tried to prove the theorem using the morphisms $mathcal{F}'(U)tomathcal{F}''(U)$ 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 $f:mathcal{F}'(U)tomathcal{F}''(U)$ and $g:mathcal{F}''(U)to mathcal{F}'(U)$. Then, by commutativity we can decompose $rho_{UV}:mathcal{F}'(U)tomathcal{F}(V)$ as $mathcal{F}'(U)to mathcal{F}''(U)tomathcal{F}'(U)to mathcal{F}(V)$, which is $fgcircrho_{UV}$. Therefore we have $rho_{UV}circ id_V=fgcircrho_{UV}$ for every $V$. Using the hypothesis, $fg$ must be the identity, and therefore we have the isomorphism.