# Understanding the inverse limit and universal property of topological spaces

Let $$\{X_i, \varphi_{ij},I\}$$ be an inverse system of topological space index by a directed poset $$I$$. Now I would like to understand the proof for the existence of an inverse limit $$(X,\varphi_i)$$.

We define $$X = \left\{ (x_i)\in \prod_{i \in I} X_i \: \middle|\: \varphi_{ij}(x_i) = x_i \text{ whenever } i \succeq j \right\}$$ and $$\varphi_i: X \to X_i$$ be the restriction of the canonical projection $$\pi_i: \prod_{j \in I} X_j \to X_i$$ on $$X$$.

• Since the $$\pi_i$$ are continuous by the definition of the product topology, the $$\varphi_i$$ must be continous too (as restrictions of continous maps are continuous).
• By construction, the maps $$\varphi_i$$ are compatible, i.e. for $$i \succeq j$$ we have $$\varphi_{ij} \circ \varphi_i = \varphi_j$$.

Now I need to show that the universal property is satisfied:

Let $$Y$$ be a topological space and $$\psi_i: Y \to X_i$$ be a set of compatible continuous mappings.

• We can define a map $$\psi:Y \to X$$ as follows: If $$y \in Y$$, then define $$\psi(y) = x$$ where is the element $$x = (x_i) \in X$$ defined by $$x_i = \psi_i(y)$$ for each $$i \in I$$. By construction, we have $$\varphi_i \circ \psi = \psi_i$$. But but I did not understand yet why $$\psi$$ is continuous:

Let $$O \subseteq X$$ be an open subset. Then I need to show that $$\psi^{-1}(O)$$ is open too. If only we could we find a way to describe $$O$$ with $$\varphi_i^{-1}(O_i)$$ for open sets $$O_i \subseteq X_i$$, then that would be nice because then $$\psi^{-1} ( \varphi_i^{-1}(O_i) ) = \psi_i^{-1}(O_i)$$ which is open because $$\psi_i$$ is continous by assumption.

• I believe this follows from the universal property of the product space $\prod_{i \in I} X_i$, of which $X$ is a subspace. – Dan Rust Feb 3 '19 at 0:16

This follows from the UMP of the product topology, which in turn, follows from a more general fact: let $$(Z,\tau)$$ be a topological space, $$X$$ a space, and $$\mathscr F=\{\phi_i:X\to Y;\ i\in I\}.$$ Now, give $$X$$ the weak topology $$(\sigma(X), (\phi_i)_{i\in I})$$ induced by $$\mathscr F$$, and finally, let $$f : Z \to X$$ be a map.
Then, $$f$$ is continuous for $$\tau$$ and $$(\sigma(X), (\phi_i)_{i\in I})$$ if and only if for every $$i \in I,\ \phi_i\circ f$$ is continuous:
$$(\Rightarrow):$$ suppose $$V$$ is open in $$Y$$. Then, $$\phi_i^{-1}(V)$$ is open in $$X$$ (by definition of the weak topology), so $$(\phi_i\circ f)^{-1}(V)=f^{-1}(\phi_i^{-1}(V))$$ is open in $$Z$$ because $$f$$ is continuous.
$$(\Leftarrow):$$ suppose $$\phi_i\circ f$$ is continuous and let $$U$$ be open in $$X$$. Then, $$U$$ is a union of finite intersections of sets of the form $$\{\phi_i^{-1}(V):i\in I;\ V\in \tau_Y\}$$ so it suffices to consider an arbitrary $$\textit{subbasis}$$ element, $$\phi_i^{-1}(V).$$ In this case, we find that $$f^{-1}(U)=f^{-1}(\phi_i^{-1}(V))=(\phi_i\circ f)^{-1}(V)$$ is open in $$Z$$ so $$f$$ is continuous.
• If you take a space $X$ and a set of maps $(f_i)$ from $X$ to a topological space $Y$, you can topologize $X$ by declaring the open sets to be unions of finite intersections of sets $f^{ -1}(V)$ with $V$ open in $Y$. This topology is the weak topology generated by the functions $(f_i)$, exactly how you define the product topology: $f_i$ are just the projections from the product to the component spaces. If you go through my answer, identifying $X,Y,Z$ and $\phi_i$ , you'll see how this works. I gave a more general proof because it covers a lot of other constructions and it's worth knowing, IMO. – Matematleta Feb 3 '19 at 3:20