How can I think about a morphism of locally ringed spaces?

Question

A locally ringed space is a pair $(X,\mathcal{O}_X)$ of a topological space $X$ and a sheaf of rings $\mathcal{O}_X$. Then we say that $(f,f^b):(X,\mathcal{O}_X)\rightarrow (Y,\mathcal{O}_Y)$ is a morphism of locally ringed spaces if $f:X\rightarrow Y$ is a continuous map and $f^b:\mathcal{O}_Y\rightarrow f_*\mathcal{O}_X$ is a morphism of sheafs on $Y$ s.t. the induced map for $x\in X$ $$f^b_x:\mathcal{O}_{Y, f(x)}\rightarrow \left(f_*\mathcal{O}_X\right)_{f(x)}\rightarrow O_{X,x}$$ satisfies $$f(\mathfrak{m}_{f(x)})\subseteq \mathfrak{m}_x$$where $\mathfrak{m}_{f(x)}$ is the maximal ideal of $\mathcal{O}_{Y, f(x)}$

I somehow understand this definition up to the point where we start speaking about this induced map. I don't see what the mean by $\mathcal{O}_{Y,f(x)}$ or all the other double indices there and I also don't see why we have this extra condition or more precisely how this extra condition works. Do we really need this condition for the definition of a morphism of locally ringed spaces?

Can someone explain this to me?

Answer 1

One reason for us to work with locally ringed spaces is that a scheme is a locally ringed space.

The condition that $f_x^b$ is a local ring morphism implies that when $X$ and $Y$ are two schemes, whenever you have two open affine subschemes $SpecA \subset X$ and $SpecB \subset Y$ such that $f$ maps $SpecA$ into $SpecB$, $f: SpecA \to Spec B$ is induced by some ring morphism $\phi: B \to A$.

For your question on stalks, you can unpack the definition of colimit and see that since we are working with sheaves of rings, elements of a stalk $\mathcal{O}_{X,x}$ are "germs", i.e. equivalence classes $[f,U]$ where $f \in \mathcal{O}_X(U)$ and U is open and contains x. [f,U]=[g,V] whenever f and g, when restricted to $U\cap V$, are the same.

Answer 2

Think of $O_X$ as a sheaf of "good" functions on the space, and $m_p \subseteq O_{X, p}$ is the set of all functions that have value $0$ at $p$. If $f \in O_{X, p}$ does not have value $0$, then $1/f$ is a well-defined function near $p$, so $f$ should be invertible in $O_{X, p}$. This is why it makes sense for $O_{X, p}$ to be a local ring and $m_p$ to be its maximal ideal.

This is literally true if we look at some examples of locally ringed spaces. For example, if $O_X$ is the sheaf of continuous functions $X \rightarrow \mathbb{R}$, then $O_{X, p}$ is a local ring and the maximal ideal is the functions that are zero at $p$.

For a morphism $(f, f^b) : X \rightarrow Y$, we can think of $f^b$ as "pulling back" functions on $Y$ to functions on $X$. For example, if $f: X \rightarrow Y$ is any continuous function, and $U \subseteq Y$ is open, and $g : U \rightarrow \mathbb{R}$ is continuous (so $g \in O_Y(U)$), then $g \circ f : f^{-1}(U) \rightarrow \mathbb{R}$ is a continuous function , so $g \circ f \in O_X(f^{-1}(U))$.

Of course, if $x \in X$ and $y = f(x)$, and $g(y) = 0$, then $(g \circ f)(x) = 0$ as well, so the pullback should send a function vanishing at $f(x)$ to a function vanishing at $x$. This is what $f^b_x(m_y) \subseteq m_x$ means.

We can do the same for smooth functions on a smooth manifold, holomorphic functions a complex manifold, and in algebraic geometry, on a scheme as well.

The difference with a scheme is that elements of $O_Y(U)$ are not really "functions" in the intuitive sense. Therefore just by having a continuous function $f$, we don't immediately know how to "pull back" elements of $O_Y(U)$ to $O_X(f^{-1}(U))$. This is why we need to explicitly include $f^b$ in the definition of a morphism of locally ringed spaces.