Uniqueness of a homomorphism in a projective limit dynamical system; showing that a projective limit system is compact - Is my proof correct? $\newcommand{\A}{\mathcal{A}}$In all that follows, $(X;\phi)$ is said to be a topological system if $X$ is a nonempty compact Hausdorff space and $\phi:X\to X$ is a continuous map. A homomorphism of topological systems $\Psi:(X_1;\phi_1)\to(X_2;\phi_2)$ is a continuous map satisfying:
$$\Psi\circ\phi_1=\phi_2\circ\Psi$$
From this text, exercise $18$, pages $29-30$:

Let $(I,\le)$ be a nonempty directed set, and let for each $i\in I$ a topological system $(K_i;\phi_i)$ be given. Suppose that for each pair $(i,j)\in I^2,\,i\le j$, a homomorphism $\pi_{ij}:(K_j;\phi_j)\to(K_i;\phi_i)$ is given, subject to the relations: $$\pi_{ii}=\mathrm{id},\quad\pi_{ij}\circ\pi_{jk}=\pi_{ik},\quad i\le j\le k\in I$$
Let: $$K=\{(x_i)_{i\in I}\in\prod_{i\in I}K_i:\pi_{ij}(x_j)=x_i,\quad i\le j\in I\}$$
And let $\pi_i:K\to K_i$ be the $i$th natural projection. By construction $\pi_{ij}\pi_j=\pi_i,\,i\le j$.
When endowed with the product dynamic $\phi=(\phi_i)_{i\in I}$, the topological system $(K;\phi)$ is called the projective limit system of $((K_i;\phi_i),\pi_{ij})$.

*

*Show that $K$ is a nonempty compact space, under the product topology, invariant under $\phi$ - that is, show $K$ is a legitimate topological system.

*Whenever $(L;\psi)$ is a topological system and $\sigma_i:(L;\psi)\to(K_i;\phi_i)$ is a homomorphism for all $i\in I$ with the added condition $\sigma_i=\pi_{ij}\sigma_j,\,i\le j$ (I refer to this as the projective property), then show that there is a unique homomorphism $\tau:(L;\psi)\to(K;\phi)$ with $\pi_i\circ\tau=\sigma_i$ for all $i\in I$.


This is more or less verbatim. Firstly, I will present a solution to $1)$ and would like verification. Secondly, I'd like some help understanding the phrasing of $2)$: do they want me to show:

$2i)$: If the $(\sigma_i)_{i\in I}$ exist as stated, then any homomorphism $\tau:L\to K$ must satisfy $\pi_i\tau=\sigma_i$ for all $i\in I$, that is: $\tau$ exists uniquely.

Or do they want me to show:

$2ii)$: There is one and only one homomorphism $\tau:L\to K$ satisfying $\pi_i\tau=\sigma_i$.

Statement $2ii$ is essentially trivial, whereas statement $2i$ is  very strong, asserting that the $\sigma_i$ uniquely define $\tau$ even though the $\sigma_i$ are arbitrary and there could be many different families of homomorphisms. Then statement $2i$ seems incorrect, but statement $2ii$ seems ridiculous since $\pi_i\tau=\sigma_i$ would imply: $$\tau:x\mapsto(\sigma_i x)_{i\in I}$$Which immediately defines $\tau$ uniquely w.r.t the $\sigma_i$. It just remains to show that this is indeed a homomorphism.
Proposed solution to $1)$:

$I$ is nonempty, so $\exists\alpha\in I$. Each of the $K_i$ are nonempty compact Hausdorff spaces. Consider: $$\tag{$\ast$}(\pi_{\alpha i}K_i)_{I\ni i\ge\alpha}$$If $i'\ge i\ge\alpha$, then: $$\pi_{\alpha i'}K_{i'}=\pi_{\alpha i}\pi_{i i'}K_{i'}\subseteq\pi_{\alpha i}K_i$$ Then $(\ast)$ is a decreasing chain of compact sets, since the homomorphisms are continuous and therefore map the compact $K_i$ to compact sets. By Cantor's Intersection Theorem: $$K_{\alpha}\supseteq\bigcap_{i\ge\alpha}\pi_{\alpha i}K_i=\A\neq\varnothing$$Moreover, as $K_\alpha$ is a Hausdorff space, and each of the images $\pi_{\alpha i}K_i$ is compact, we have that the images $\pi_{\alpha i}K_i$ are all closed subsets in $K_\alpha$, and so their intersection $\A$ is closed. Then $\A$ is a nonempty compact set.
We can then choose an $x_\alpha\in\A$. For all $i\le\alpha$, set $x_i=\pi_{i\alpha}x_\alpha$. Let $i\ge\alpha$ be arbitrary: by choice of $x_\alpha$ the preimage $\pi_{\alpha i}^{-1}\{x_\alpha\}$ is nonempty for all $i\ge\alpha$, and as such one can choose arbitrarily an $x_i\in\pi_{\alpha i}^{-1}\{x_\alpha\}$ so that $\pi_{\alpha i}x_i=x_\alpha$. Then for $\alpha\le i'\le i$, set $x_{i'}=\pi_{i'i}x_i$. Then choose a $j\ge i$, and select $x_j\in\pi_{\alpha j}^{-1}\{x_\alpha\}$, and put for all $i\le j'\le j$ $x_{j'}=\pi_{j' j}x_j$. Perform this ascent throughout $I$.
With all the $x_i$ chosen in this manner, put $(x_i)_{i\in I}$ which definitely exists in $\prod_{i\in I}K_i$. It remains to check the property $\pi_{ij}x_j=x_i,\,\forall i\le j$. By construction, we can say that: $$\begin{align}&i\le j\le\alpha:\quad \pi_{ij}x_j=\pi_{ij}\pi_{j\alpha}x_\alpha=\pi_{i\alpha}x_\alpha=x_i\\&j\ge\alpha\ge i:\quad\pi_{ij}x_j=\pi_{i\alpha}\pi_{\alpha j}x_j=\pi_{i\alpha}x_\alpha=x_i\\&j\ge i\ge\alpha:\quad\pi_{ij}x_j=x_i\end{align}$$Therefore $(x_i)_{i\in I}$ exists and it exists in $K$, so $K$ is non-empty.
By the exact same argument, $\exists\alpha\in I$ and any $x\in K$ must satisfy $x_\alpha\in\A$, where $\A$ is constructed as before. That is, $x\in K\implies\pi_\alpha x\in\A$, and so $\pi_\alpha(K)=\A$. As discussed, $\A$ is closed and compact, and under the product topology the projection $\pi_\alpha$ is continuous, so $K$ must be compact.
Finally, consider $\phi(x)$ for $x\in K$. $$\phi(x)=(\phi_i\pi_i x)_{i\in I}$$We want to check if $\pi_{ij}(\phi(x))_{j}=(\phi(x))_{i},\,i\le j$. $$\pi_{ij}(\phi(x))_{j}=\pi_{ij}\phi_j\pi_jx\overset{\text{Hom}}{=}\phi_i\pi_{ij}\pi_j x=\phi_i\pi_i x=(\phi(x))_i$$Recalling that the $\pi_{ij}$ are homomorphisms. Therefore $\phi(x)\in K,\,\forall x\in K,\,\phi(K)\subseteq K$, and $K$ is $\phi$-invariant.

Now, about $2)$: the best that I have done is show that if $\pi_\alpha\tau=\sigma_\alpha$ for at least one $\alpha\in I$, then $\pi_i\tau=\sigma_i$ for all $i\in I$. I am left troubled with the notion of $\tau$ being "unique", since the $\sigma_i$ are arbitrary homomorphisms.

Assume $\exists\alpha\in I:\pi_\alpha\tau=\sigma_\alpha$. For any $i\in I$: $$\tau\psi=\phi\tau=(\phi_i\pi_i\tau)_{i\in I}\\\pi_i\tau\psi=\phi_i\pi_i\tau=\phi_i\pi_{ij}\pi_j\tau=\pi_{ij}\phi_j\pi_j\tau$$For any $j\ge i$.
Let $i\le\alpha$ and $j=\alpha$ in the above equation: $$\pi_i\tau\psi=\pi_{i\alpha}\phi_\alpha\pi_\alpha\tau=\pi_{i\alpha}\phi_\alpha\sigma_\alpha=\pi_{i\alpha}\sigma_\alpha\psi=\sigma_i\psi\implies\pi_i\tau=\sigma_i$$
Now suppose $i\ge\alpha$ and reverse the roles: $$\pi_{\alpha i}\phi_i\pi_i\tau=\pi_\alpha\tau\psi=\sigma_\alpha\psi=\pi_{\alpha i}\sigma_i\psi=\pi_{\alpha i}\phi_i\sigma_i\implies\pi_i\tau=\sigma_i$$
Therefore $\exists\alpha\in I:\pi_\alpha\tau=\sigma_\alpha$, whenever $\tau,(\sigma_i)$ are homomorphisms with the stated properties, implies that $\pi_i\tau=\sigma_i,\,\forall i\in I$.  That is, $\tau$ is uniquely determined by this family $\sigma_i$ if it exists and coincides in at least one place.
Let $\lambda_i=\pi_i\tau$, where $\tau:L\to K$ is any arbitrary homomorphism. $$\lambda_i\psi=\pi_i\tau\psi=\pi_i\phi\tau=\phi_i\pi_i\tau=\phi_i\lambda_i$$So it follows that $(\lambda_i)_{i\in I}$ is a family of homomorphisms, $\lambda_i:L\to K_i$. Let $i\le j$: $$\pi_{ij}\lambda_j=\pi_{ij}\pi_j\tau=\pi_i\tau=\lambda_i$$So $(\lambda_i)_{i\in I}$ satisfies the asserted conditions on $(\sigma_i)_{i\in I}$. Then any homomorphism $\tau:L\to K$ uniquely defines a family of homomorphisms $(\sigma_i)_{i\in I}$ with the projective property, and any family of homomorphisms $(\sigma_i)_{i\in I}$ with the projective property uniquely define a homomorphism $\tau$ by setting $\pi_\alpha\tau=\sigma_\alpha$ for one $\alpha\in I$.

Am I done? There seems to be an issue with uniqueness, since if I have a family $(\bar{\sigma_i})_{i\in I}$ of homomorphisms $L\to K_i$ with the projective property, then I can immediately define a different homomorphism $\bar{\tau}:L\to K$. If I am given a homomorphism $\tau:L\to K$, I can uniquely identify $\tau$ with a family $(\sigma_i)_{i\in I}$ of homomorphisms with the projective property by simply setting $\sigma_\alpha:=\pi_\alpha\tau$ for one $\alpha\in I$. Is this what the question was asking?
 A: Your solution to 1) seems to be essentially correct.
Regarding 2), your interpretation 2ii) is the correct one. More specifically, there may be many homomorphisms from an anonymous $(L,\psi)$ to the inverse limit system $(K,\phi)$, but a $\sigma_\bullet:(L,\psi)\to (K_\bullet,\phi_\bullet)$ ($\bullet$ signifies the whole family) determines a unique homomorphism $\tau: (L,\psi)\to(K,\phi)$ (which might as well be denoted by $\sigma$ according to the conventions of notation of the exercise in question). As you said if $\sigma_\bullet : L\to K_\bullet$ exists, since $K$ is defined as a certain subset of $\prod K_\bullet$, there is a unique arrow $\tau : L\to \prod K_\bullet$, whose coordinates are given by $\sigma_\bullet$. If in addition $\sigma_\bullet$ is compatible with the projections $\pi_{\bullet\bullet}$, then the image of $\tau$ is a subset of $K$, and the fact that $\sigma_\bullet$ is a homomorphism $\psi\to \phi_\bullet$ gives that it's a homomorphism $\psi\to \phi$.
Conversely, if $\tau:(L,\psi)\to (K,\phi)$ is an anonymous homomorphism, then we have $\sigma_\bullet:=\pi_\bullet\circ \tau: (L,\psi)\to (K_\bullet,\phi_\bullet)$ compatible with $\pi_{\bullet\bullet}$ and based on the previous paragraph if $\tau': (L,\psi)\to (K,\phi)$ is an anonymous homomorphism with $\pi_\bullet\circ \tau'=\pi_\bullet\circ \tau$, then $\tau'=\tau$. For practical purposes all this means that when dealing with homomorphisms to/from the inverse limit system you may do so component-wise.
Regarding your use of "ridiculous", the point of category theory (when applicable) is to showcase the non-arbitrariness of constructions that might seem otherwise (see e.g. "The purpose of being categorical is to make that which is formal, formally formal" what does it mean?).
