An algebraic geometry assignment asks the reader to show that $\mathbb{P}^1(\mathbb{C})$ is two copies of $\mathbb{C}$ "glued together." The meaning of "glued together" isn't defined but the emphasis in this subject is definitely on ringed spaces; the lecture notes define the closed subsets of $\mathbb{P}^n$ as vanishing sets of homogeneous polynomials, and then proceed to define the structure sheaf.
I was thinking that a good way to proceed is this:
Firstly, let $\alpha : \mathbb{C}_{\neq 0} \rightarrow \mathbb{C}$ be given by $\alpha_0(z) = z$ and let $\alpha_1: \mathbb{C}_{\neq 0} \rightarrow \mathbb{C}$ be given by $\alpha_1(z) = 1/z.$ The claim to be proved is that the pushout of $\alpha_0$ and $\alpha_1$ is $\mathbb{P}^1$ in the category of locally ringed spaces.
More precisely, define $\beta_0 : \mathbb{C} \rightarrow \mathbb{P}^1$ and $\beta_1 : \mathbb{C} \rightarrow \mathbb{P}^1$ by $\beta_0(z) = [z:1]$ and $\beta_1(z) = [1:z]$ respectively. Proceed by showing that $(\beta_0,\beta_1)$ is the pushout of $(\alpha_0,\alpha_1)$ in the category of locally ringed spaces.
We need to show that:
- $(\beta_0,\beta_1)$ is a cocone from $(\alpha_0,\alpha_1)$
- $(\beta_0,\beta_1)$ is initial in the category of cocones from $(\alpha_0,\alpha_1)$
Part 1 is easy. We have to show that $\beta_0 \circ \alpha_0 = \beta_1 \circ \alpha_1$, which amounts to proving that $[z:1] = [1:1/z]$ for any non-zero complex number $z$. But this is trivial.
Part 2 is harder. I was thinking there might be a high-powered tool available the can help us; in particular, I was wondering if there's a tool that could allow us to deduce the initiality of $(\beta_0,\beta_1)$ by verifying some simpler facts, like the fact that $\mathrm{img}(\beta_0) \cup \mathrm{img}(\beta_1) = \mathbb{P}^1$ (which is easy to show).
Ideas anyone?
Not so much high-powered, just how colimits of so-called glueing data work. The details can be found here: stacks.math.columbia.edu/tag/01JA. To answer the particular question here, however, we are not actually concerned with the existence (which is proved in loc. cit.), but the uniqueness (whose proof is omitted).
Since the argument is literally the same in either the category of ringed, or locally ringed spaces, I will simply say space. Recall that a morphism of spaces $f\colon (X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ is given by a continuous map $f\colon X\to Y$ and a certain morphism $\mathcal O_Y\to f_*\mathcal O_X$, or, by adjunction, $f^{-1}\mathcal O_Y\to\mathcal O_X$. In particular, if $U\subset X$ is open, then the restriction of the latter morphism to $U$ gives $(f|_U)^{-1}\mathcal O_Y = f^{-1}\mathcal O_Y|_U\to\mathcal O_X|_U = \mathcal O_U$, which corresponds to the restriction of $f$ to $(U,\mathcal O_U)$. In particular, if $X = \bigcup_{i\in I}U_i$, then we get a bunch of morphisms $f_i := f|_{U_i}\colon (U_i,\mathcal O_{U_i})\to (Y,\mathcal O_Y)$. Note that these satisfy $f_i|_{U_i\cap U_j} = f_j|_{U_i\cap U_j}$ for all $i,j\in I$, where $=$ is meant as morphisms of spaces, including the sheaf morphisms.
I claim that this defines a natural bijection $$\hom((X,\mathcal O_X),(Y,\mathcal O_Y)) = \left\{(f_i)_i\in\prod_{i\in I}\hom((U_i,\mathcal O_{U_i}), (Y,\mathcal O_Y))\middle|\;\forall i,j\in I,\;f_i|_{U_i\cap U_j} = f_j|_{U_i\cap U_j}\right\}$$ In other words, what I claim is that given $(f_i)_i\in\prod_{i\in I}\hom((U_i,\mathcal O_{U_i}), (Y,\mathcal O_Y))$ satisfying $f_i|_{U_i\cap U_j} = f_j|_{U_i\cap U_j}$ for all $i,j\in I$, there exists a unique $f$ such that $f|_{U_i} = f_i$. (Again, all $=$ meaning as morphisms of spaces.) On the level of topological spaces, this is easy; the existence and uniqueness of the map of underlying sets is trivial and that it is continuous follows from the local characterisation of continuity. Thus, we have to show that the sheaf morphisms $f^{-1}\mathcal O_Y|_{U_i} = f_i^{-1}\mathcal{O}_Y\to\mathcal{O}_{U_i} = \mathcal{O}_X|_{U_i}$ glue uniquely to a morphism of sheaves $f^{-1}\mathcal O_Y\to\mathcal O_X$. But by the sheaf property, sheaf morphisms do indeed glue uniquely from an open cover as soon as they agree on pair-wise overlaps, which is exactly what we assumed.
Now, what has this got to do with the original question? Well, this shows in particular that if a space, say $\mathbb P^1(\mathbb C)$, is the union of two open sub-spaces $U_1\cup U_2 = \mathbb P^1(\mathbb C)$, then it is the push-out of the span $U_1\leftarrow U_1\cap U_2\to U_2$. In particular, if we have spaces $X_1$, $X_2$, $X_{12}$, and a span $X_1\leftarrow X_{12}\to X_2$, where both morphisms are isomorphisms onto open sub-spaces, and isomorphisms $X_i\to U_i$ such that $X_{12}$ is isomorphically mapped onto $U_1\cap U_2$, then the two spans are isomorphic; hence, their push-outs $X_1\amalg_{X_{12}}X_2$ and $\mathbb{P}^1(\mathbb{C})$ are naturally isomorphic as well.