In studying deformation theory of Galois representations, I've come surely to an error, relating Schlessinger's criterion. 


Let's fix a representation $\bar{\rho}$ of a group $G$ and let $D\_{\bar{\rho}}$ be its deformation functor from the category $\hat{\mathcal{C}}$ of complete noetherian local $W(\mathbb{F})$-algebras (where $W(\mathbb{F})$ is the ring of the Witt vectors over a finite field $\mathbb{F}$) with residue field $\mathbb{F}$. Let $A'$, $A''$ and $A$ be artinian objects of $\hat{\mathcal{C}}$, and let be given two homomorphisms of algebras $u' : A' \rightarrow A$ and $u'': A'' \rightarrow A$. One of the conditions of Schlessinger's criterion, namely that the natural map $$D\_{\bar{\rho}}(A'\times\_A A'') \rightarrow D\_{\bar{\rho}}(A') \times\_{D\_{\bar{\rho}}(A)} D\_{\bar{\rho}}(A'')$$ should be surjective when $u'$ is a small surjection, seems to me always true, regardless the surjectivity of $u'$. Take in fact $([\rho'], [\rho'']) \in D\_{\bar{\rho}}(A') \times\_{D\_{\bar{\rho}}(A)} D\_{\bar{\rho}}(A'')$ and let $\rho: G \rightarrow \operatorname{GL}\_n(A'\times\_A A'')$ be defined as $\rho(\sigma) = (\rho'(\sigma), \rho''(\sigma))$, using the identification $$\operatorname{GL}\_n(A'\times\_A A'') \simeq \operatorname{GL}\_n(A')\times\_{\operatorname{GL}\_n(A)}\operatorname{GL}\_n(A'').$$ By the property of this last identification, we see that by conjugating one of $\rho'$ or $\rho''$, we conjugate also $\rho$. Thus this yields a well defined $[\rho] \in D\_{\bar{\rho}}(A'\times\_A A'')$ and seems just the one we wanted.


Since this must clearly be wrong, where is my mistake?


Sorry if this question reveals my big miscomprehension! Thank you in advance.



Misunderstanding in the hypotheses of Schlessinger's criterion
I think the "mistake" is in the definition of the equivalence relation that defines a representation. Indeed, an $A$-valued representation (for some $A\in\hat{C}$) is a *conjugacy class* of homomorphisms $$
\rho:G\to \operatorname{GL}\_n(A)
$$
reducing to $\bar{\rho}$. This makes the equality
$$
D\_\bar{\rho}(A'\times\_AA'')=D\_\bar{\rho}(A')\times\_{D\_\bar{\rho}(A)}D\_{\bar{\rho}}(A'')
$$
false, in general, because the sets $D\_\bar{\rho}(\cdot)$ are smaller than you would expect (and this can make the fiber product behave oddly). The point is that two homomorphisms $\rho'\colon G\to\operatorname{GL}\_n(A')$ and $\rho''\colon G\to\operatorname{GL}\_n(A'')$ might reduce to different morphisms in $\operatorname{GL}\_n(A)$ (so, the pair does not define an element in the left-hand side) but these reductions might be conjugate in $\mathrm{GL}\_n(A)$ thus belonging to the right-hand side. Surjectivity of $u\colon A'\to A$ helps you out of this trouble.


**A word of warning**: actually, what I wrote above is slightly incorrect. It would be correct if we were considering the functor assigning to each $A$ the set of $A$-valued representations of $G$ with values in $\operatorname{GL}\_n(A)$; but the functor $D\_\bar{\rho}$ one normally considers while deforming Galois representations is a bit different, namely you are allowed to conjugate only by matrices in $$
\operatorname{Ker}\Big(\operatorname{GL}\_n(A)\to\mathrm{GL}\_n(\mathbb{F})\Big).
$$
In other words, if $\rho,\rho'$ verify $M\rho(g)M^{-1}=\rho'(g)$ for some $M\in\operatorname{GL}\_n(A)$ and all $g\in G$, then the *representations* $\rho,\rho'$ are isomorphic but they define the same *deformation* only if $M\equiv \mathrm{id}.\pmod{m\_A}$ where $m\_A$ is the maximal ideal of $A$. In particular, the set $D\_\bar{\rho}(A)$ is even bigger than before.


You find a detailed proof that surjectivity of $u'$ implies what you want in the right language of deformations (instead of mere representations) for instance in Section 3.1 of Tilouine's *Deformations of Galois Representations and Hecke Algebras*, Mehta Res. Institute (1995); on page 391 of his original Deformations paper in *Galois groups over $\mathbb{Q}$* Mazur gives an extremely succinted discussion - actually, he just says "it is easy", so it might be of small use.


