= Solution
For the <cofinite topology> $\tau_1$, the empty set and $\mathbb N$ are included. A nonempty union of cofinite sets is cofinite because its complement is contained in the finite complement of any one member. A finite intersection is cofinite because its complement is a finite union of finite sets; intersections involving the empty set are empty. This proves all <topology> axioms.
For the <tail topology on the natural numbers> $\tau_2$, $\mathbb N=I_1$. Any nonempty collection of tails has a smallest starting index, so its union is that tail; a finite intersection of tails is the tail with largest starting index. Empty sets and empty-family conventions supply the remaining cases. Thus $\tau_2$ is also a <topology>.
We classify <continuous maps between cofinite and tail topologies>. <Continuity> is equivalent to
$$
E_k=\{n:f(n)\ge k\}\text{ being empty or cofinite for every }k.
$$
If $f$ has unbounded image, each $E_k$ is nonempty, hence cofinite. For every threshold $k$, eventually $f(n)\ge k$, which is exactly $f(n)\to\infty$. If the image is bounded, it is a nonempty finite subset of $\mathbb N$ and has a maximum $N$. Then $E_N$ is nonempty and cofinite, but $f(n)\le N$ always, so $f(n)=N$ except at finitely many indices. These exhaust the two possibilities and prove necessity. Their sufficiency is checked separately below.
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