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Let $X$ be a differentiable manifold. Let $\mathcal{O}_X$ be the sheaf of $\mathcal{C}^\infty$ functions on $X$. Since every stalk of $\mathcal{O}_X$ is a local ring, $(X, \mathcal{O}_X)$ is a locally ringed space. Let $Y$ be another differentiable manifold. Let $f\colon X \rightarrow Y$ be a differentiable map. Let $U$ be an open subset of $Y$. For $h \in \Gamma(\mathcal{O}_Y, U)$, $h\circ f \in \Gamma(\mathcal{O}_X, f^{-1}(U))$. Hence we get an $\mathbb{R}$-morphism $\Gamma(\mathcal{O}_Y, U) \rightarrow \Gamma(\mathcal{O}_X, f^{-1}(U))$ of $\mathbb{R}$-algebras. Hence we get a morphism $f^{\#} \colon \mathcal{O}_Y \rightarrow f_*(\mathcal{O}_X)$ of sheaves of $\mathbb{R}$-algebras. It is easy to see that $(f, f^{\#})$ is a morphism of locally ringed spaces.

Conversely suppose $(f, \psi)\colon X \rightarrow Y$ is a morphism of locally ringed spaces, where $X$ and $Y$ are differentiable manifolds and $\psi\colon \mathcal{O}_Y \rightarrow f_*(\mathcal{O}_X)$is a morphism of sheaves of $\mathbb{R}$-algebras. Is $f$ a differentiable map and $\psi = f^{\#}$?

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    $\begingroup$ Yes. The point is that once you have a morphism of ringed spaces then you know that the map has an expression in local coordinates that is smooth/analytic/algebraic etc. as according to the nature of your structure sheaf. Brian Conrad has notes on the locally ringed space approach to differential geometry, if I recall correctly. $\endgroup$
    – Zhen Lin
    Commented Oct 30, 2012 at 22:08
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    $\begingroup$ @ZhenLin: Could you provide a link to these notes (or notes [or even books] treating differential geometry also in the language of locally ringed spaces)? Unfortunately I can't find them. $\endgroup$
    – Hanno
    Commented Feb 21, 2015 at 7:40
  • $\begingroup$ I misremembered. It was Keith Conrad, not Brian. See here. $\endgroup$
    – Zhen Lin
    Commented Feb 21, 2015 at 7:57
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    $\begingroup$ @ZhenLin I'm very interested in such notes! I looked at the link you gave, and I didn't see anything on locally ringed spaces in there. Also, those notes are from Brian Conrad, not Keith. Did you maybe post the wrong link? $\endgroup$
    – Eric Auld
    Commented Apr 25, 2016 at 18:00
  • $\begingroup$ Perhaps I mean here specifically. $\endgroup$
    – Zhen Lin
    Commented Apr 25, 2016 at 18:52

2 Answers 2

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Yes: Let $(f,\psi):X\to Y$ be a morphism of locally ringed spaces, where $X$ and $Y$ are smooth manifolds with their sheaves of smooth functions. If $\psi:C^\infty_Y \to f_* C^\infty_X$ is a morphism of sheaves of $\mathbb R$-algebras, then $f$ is smooth and $\psi=f^\#$.

Proof. Let $s:U\to \mathbb R$ be a smooth function. The equation $\psi s= s\circ f$ follows from the commutativity of the diagram below. Notice the triangle commutes because there is a unique $\mathbb R$-algebra map $C^\infty_{f(x)}/{\frak m}_{f(x)}\cong \mathbb R \to \mathbb R$. It now follows that $f:X\to Y$ is smooth. Indeed, we know $s\circ f$ is smooth for all real valued functions $s$ on $Y$, and we may take $s$ to be the coordinate functions of charts on $Y$. QED.

enter image description here

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    $\begingroup$ Why is $s \circ f$ smooth here? We know that $s$ is smooth but we only know that $f$ is continuous, right? $\endgroup$
    – HarshCurious
    Commented Nov 12, 2016 at 1:51
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    $\begingroup$ $s\circ f$ is smooth because, as I show above, it equals $\psi f$ (and $\psi$ is by assumption a map of sheaves of smooth functions). This observation (which holds for all smooth functions $s$ on open subsets of $Y$) implies that $f:X\to Y$, a function that a priori is only continuous, has to be smooth. $\endgroup$
    – Lucas Braune
    Commented Nov 12, 2016 at 2:03
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    $\begingroup$ Sorry, I am writing from a phone. Yes, I that's what I mean. $\endgroup$
    – Lucas Braune
    Commented Nov 12, 2016 at 2:29
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    $\begingroup$ I think the assumption that the components of the morphisms are R-algebra homomorphisms is unnecessary. Any ring endomorphism of the real numbers preserves the ordering ($a\leq b\iff b-a=c^2\implies f(b)-f(a)=f(c)^2\iff f(a)\leq f(b)$) and so is the identity (the rationals are order-dense in the reals). $\endgroup$
    – Vladimir Sotirov
    Commented Mar 28, 2022 at 17:43
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    $\begingroup$ @ElíasGuisadoVillalgordo I claimed only that ring homomorphisms between the reals, not arbitrary real algebras, are real-linear. What's happening in general is that rings of real-valued functions on a set $X$ can be identified with subrings of $\prod_{x\in X}\mathbb R\cong[X,\mathbb R]$. The argument above shows that the component of a morphism $(f,\phi)$ of locally ringed spaces is the restriction of $f^*\colon[Y,\mathbb R]\to[X,\mathbb R]$ to the subrings. But $f^*$ is an $\matbb R$-algebra homomorphism, so likewise is its restriction to an $\mathbb R$-subalgebra. $\endgroup$
    – Vladimir Sotirov
    Commented Mar 21, 2023 at 19:01
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See Prop 4.18 of the great book Manifolds, sheaves, and cohomology by Torsten Wedhorn. This book treat manifolds as locally ringed spaces.

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