Suppose that $(X,\varphi_i)$ is an inverse limit of an inverse system of compact hausdorff spaces $\{X_i,\varphi_{ij}\}$. Let $Y$ be a subspace of $X$, I would like to proof that $\overline{Y}$ (closure in X) is the inverse limit of $\{\varphi_i(Y),\varphi_{ij}\}$. Since $\varphi_i:Y\rightarrow \varphi_i(Y)$ is onto, the image of the inclusion $Y\rightarrow \text{lim} \ \varphi_i(Y)$ is dense. The problem is that I don't know how to show that $\text{lim} \ \varphi_i(Y)$ is closed in $X$.
Inverse limit of compact hausdorff spaces
general-topologylimits-colimits
Related Solutions
Nothing about the existence and uniqueness of the inverse limit relies on the assumption that $\varphi_{jk} \varphi_{ij} = \varphi_{ik}$. However, omitting this assumption does not actually give any greater generality. Indeed, note that given $X$ with maps $\varphi_i:X\to X_i$ satisfying $\varphi_{ij}\varphi_i=\varphi_j$ whenever $i\succeq j$, we have $$\varphi_{jk} \varphi_{ij}\varphi_i=\varphi_{jk}\varphi_j=\varphi_k=\varphi_{ik}\varphi_i$$ whenever $i\succeq j\succeq k$. In other words, the image of the map $\varphi_i$ must be contained in the subset $Y_i\subseteq X_i$ consisting of elements $x$ such that $\varphi_{jk} \varphi_{ij}(x) = \varphi_{ik}(x)$ whenever $i\succeq j\succeq k$. This means we can restrict the inverse system to the $Y_i$ instead of the $X_i$ (exercise: check that $\varphi_{ij}(Y_i)\subseteq Y_j$) without changing what an inverse limit of the system is. When we restrict to the $Y_i$, the equations $\varphi_{jk} \varphi_{ij} = \varphi_{ik}$ are true.
To put it another way, the assumption that $\varphi_{jk} \varphi_{ij} = \varphi_{ik}$ is essentially inherent in the condition $\varphi_{ij}\varphi_i=\varphi_j$ in the definition of the inverse limit. You could have elements of $X_i$ on which $\varphi_{jk} \varphi_{ij} = \varphi_{ik}$ is not true if you really wanted to for some reason, but those elements are irrelevant to the inverse limit.
(Note that the all above comments also apply to the assumption that $\varphi_{ii}$ is the identity map on $X_i$, which you omitted but is also part of the definition of an inverse system.)
From the perspective of Kevin Carlson's answer, dropping the condition $\varphi_{jk} \varphi_{ij} = \varphi_{ik}$ means that you are not really talking about a limit indexed by the poset $I$, but rather a different category (namely, the category freely generated by $I$ as a directed graph). In practice, limits indexed by that different category pretty much never come up naturally and do not have any special properties to differentiate them from arbitrary limits, so they are not discussed separately from general limits the way that inverse limits are.
No. Let $Y=\{0\}\cup\{1/n: n\in\Bbb N\}$ be a sequence convergent to zero, $X=Y\times\Bbb N$, $Z=\{0\}\times\Bbb N$, and $f:Z\to Y$, $(0,n)\mapsto 1/n$. It is easy to see that the space $X \sqcup_f Y$ is homeomorphic to the following Franklin’s space decribed in the following example from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989).
Moreover, let $q: X \sqcup Y\to X \sqcup_f Y$ be the quotient map. It is easy to see that any neighborhood of $q(0)$ is $X \sqcup_f Y$ contains a sequence of a form $\{q(n, m_n):n\ge N\}$ for some $N$, which has no limit points, so $X \sqcup_f Y$ is not locally compact.
Best Answer
This isn’t true. Consider an extremely simple directed system with only one object, with the unique $\phi$ simply being the identity map. Then applying your claim to this system, we get that every subset of every compact Hausdorff space is closed, which is clearly false.