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We are talking about sheaves of rings

$$\mathcal F:Open(X)^{op}\to Rings$$

If the image of $f$ in all stalks $\mathcal F_x$ for $x \in U$ is a unit, then $f \in \mathcal F(U)$ is a unit.

I have checked $f\in\mathcal{O}_X(X)$ is a unit $\Leftrightarrow f_x\neq 0$ for all $x\in X$

they are different questions.

My feeling about the question is negative. First, if $\phi,\psi$ agree on all stalks for every $x\in U$ then we can say they are equal since we can glue using sheaf condition. However, if $f$ is unit I feel we cannot glue to global unit element. Cannot construct a counterexample.

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  • Consider the statement $\exists v (v f = 1)$ in the internal logic. This statement amounts to saying that ${U \mid U$ open and $\exists v \in \mathcal{F}(U) (v \cdot f|_U = 1)}$ is an open cover. This follows from the stalks formulation provided. Now since inverses, if they exist, are unique, it follows that $\exists! v (vf = 1)$ holds in the internal logic. This means there actually does exist a global $v$ such that $vf = 1$ holds in the internal logic, by the interpretation of $\exists!$. And $vf = 1$ in the internal logic iff $vf = 1$. Thus, $f$ is a unit. – Mark Saving Jan 30 '22 at 20:38

1 Answers1

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You can glue the inverses of $f$ to a global element to conclude that there is $g \in \mathcal F(U)$ such that $(gf)_x = (fg)_x = 1_x$ for all $x$, thus concluding that $f$ is a unit.

If $f_x$ has an inverse, then in some vicinity $V \ni x$ there is something which I denote $g_x \in \mathcal F(V)$ such that $f|_V g_x = g_x f|_V = 1.$ Doing this for all $x$ creates an assignment $x \mapsto g_x,$ which we'd like to give rise to a global element $g \in \mathcal F(U).$ If $g_x$ is defined on $V_x$ and $g_y$ is defined on $V_y$, then on $V_x \cap V_y$ they both are inverses of $f|_{V_x \cap V_y}$. In general, if elements of a ring $g,h$ are inverses of $f$, then $g = gfh = h,$ so inverses have to coincide. This way, $\{g_x\}_{x \in U}$ satisfies the assumption of the sheaf condition and yield a global inverse of $f$.

Арсений Кряжев
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