My question essentially comes from EGA I (the Springer version), of the item (3.5.6) of the Préliminaires.
The ideia is that given a presheaf of a topological space $X$ going to a category $C$, we can express the sheaf associated to it, via the direct image of the inverse image through the identity function $1_X = I:X\to X$, give that the latter exists.
\begin{equation} \tilde{\mathcal{F}} = I_*I^{-1}(\mathcal{F}) = I^{-1}\mathcal{F} \end{equation}
Maybe I'm missing something out, but it is not obvious for me why $\tilde{\mathcal{F}}$ is a sheaf. I'll give some more details:
Firstly, the direct image through the identity function is, of course, $\mathcal{F}$ itself (justifying the second equality above). You'll see why I made the distinction later.
Secondly, the inverse image is defined as the left adjoint to the direct image, so we have a natural bi-isomorphism
\begin{align} \hom(I^{-1}\mathcal{F}, \mathcal{G}) &\xrightarrow{\cong} \hom(\mathcal{F}, I_*\mathcal{G})\ \ \ (= \hom(\mathcal{F}, \mathcal{G}) )\\ u &\mapsto u^\flat \end{align}
Using $\mathcal{G} = I^{-1}\mathcal F$ and using $u = 1_\mathcal{F}$ we have a natural morphism $\rho:\mathcal F \to \tilde{\mathcal F}$ satisfying the following unviersal property: for every $v: \mathcal F \to \mathcal G$ there exists a unique $v^\sharp: \tilde{\mathcal F} \to \mathcal G$ such that $v$ is the composite
\begin{equation} v: \mathcal F \xrightarrow{\rho} \tilde{\mathcal F} \xrightarrow{v^\sharp} \mathcal G. \end{equation}
OBS: the aplication $v \mapsto v^\sharp$ is the inverse of $u \mapsto u^\flat$.
This is the universal property that we would want the sheaf associated to $\mathcal F$ to satisfy (Well, actually if we restrict ourselves only to $\mathcal G$ that are sheaves ). However we need $\tilde{\mathcal F}$ to be a sheaf in the first place (for example, $\mathcal F$ has the same universal property, but it is not its sheafification).
Okay, so there are some equivalent ways of proving that a presheaf is a sheaf. We most certainly will need to use some of the properties of adjoint functors to prove this.
I thought I had solved this since left adjoint preserves colimits, so we would have something like this
\begin{equation} \tilde{\mathcal F}(\injlim U_\lambda) = \projlim{\tilde{\mathcal F}U_\lambda} \end{equation}
which is the sheaf condition (for cocomplete category $C$, but i would be completely satisfied with this)!
However we have that the inverse image functor is left adjoint to the direct image functor as functors between $pSh(X)$, and not (necessarily (I think)) we have that $\psi^{-1}\mathcal F$ is left adjoint to $\psi_*\mathcal F$ as functors of $Open(X)^\text{op} \to C$ (which would be enough by the above argument).
Any help on this? I guess by looking around there is another (more concrete) description of the inverse image of a sheaf, but I'm avoiding this. EGA seems only to mention this here, so I think this is doable using only the abstract nonsense definition.