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.

Could you please help me with this problem? Thank you in advance!

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    $\begingroup$ I believe this follows from the universal property of the product space $\prod_{i \in I} X_i$, of which $X$ is a subspace. $\endgroup$ – 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.

  • $\begingroup$ Thanks for your response! Could you please explain to me what this weak topology is? $\endgroup$ – Diglett Feb 3 '19 at 3:10
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    $\begingroup$ 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. $\endgroup$ – Matematleta Feb 3 '19 at 3:20
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    $\begingroup$ @Diglett they're also called "initial topologies" in topology; I explain them here. $\endgroup$ – Henno Brandsma Feb 3 '19 at 8:14

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