**Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$?**

The answer is Yes, there is such an embedding.

I will argue that if $\kappa$ is an
infinite cardinal, then there is a complete lattice embedding
$\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa\times\kappa)$.
This is enough to answer the question, for the following reason.
Any bijection $\beta:\kappa\times\kappa\to\kappa$ induces a lattice isomorphism $\overline{\beta}: \text{Top}(\kappa\times\kappa)\to \text{Top}(\kappa)$ which maps the cofinite topology on $\kappa\times\kappa$ to the cofinite topology on $\kappa$. A topology is $T_1$ iff it contains the cofinite topology, so $\overline{\beta}$ restricts to a lattice isomorphism from $\text{Top}^{T_1}(\kappa\times\kappa)$ to $\text{Top}^{T_1}(\kappa)$.
Thus, any (complete) lattice embedding $\text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa\times\kappa)$ can be altered to a (complete) lattice embedding
$\text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$ by composing with such a $\overline{\beta}$.

If $U\subseteq \kappa$, then by a
**cofinite extension of $U$** I mean a subset
$X\subseteq U\times \kappa$ where, for each $u\in U$, the set
$\{\lambda<\kappa\;|\;(u,\lambda)\in X\}$ is cofinite in $\kappa$.
To make sure this is clear, let me explain this a second way
using the projection maps $\pi_1, \pi_2\colon \kappa\times\kappa\to\kappa$.
$X$ is a cofinite extension of $U$ if
(i) $\pi_1(X)=U$, and (ii) for every $u\in U$ we have
$\pi_2((\{u\}\times\kappa)\cap X)$ is cofinite in $\kappa$.

If $\tau$ is a topology, let $\widehat{\tau}$
be the collection of all cofinite extensions
of sets in $\tau$. I claim that

**I.** For any topology $\tau$ on $\kappa$,
$\widehat{\tau}$ is a $T_1$ topology on $\kappa\times\kappa$.

**II.** The map $\tau\mapsto \widehat{\tau}$ is a
complete lattice embedding of
$\text{Top}(\kappa)$ into $\text{Top}^{T_1}(\kappa\times\kappa)$.

These are not hard to prove and they establish the result.

In the following justifications, if $X$ is a cofinite
extension of $U$, then I may refer to the *fibers*
of $X$, by which I mean fibers of $X$ under
the first projection $\pi_1$. If $x\in\pi_1(X)$,
then the fiber of $X$ over $x$ is
$(\{x\}\times\kappa)\cap X$,
which is a subset of $\kappa\times\kappa$.
(So, a cofinite extension of $U\subseteq \kappa$
is a subset if $U\times \kappa$ with cofinite fibers.)

**Sketch of proof of I.**
(Least and largest subsets)
The least and largest
subsets $\emptyset$ and $\kappa\times\kappa$ of the set
$\kappa\times\kappa$ are cofinite extensions of the
least and largest subsets
$\emptyset$ and $\kappa$ of $\kappa$.

(Finite intersection)
If $X, Y\in \widehat{\tau}$, then they are cofinite
extensions of some $\pi_1(X)=U, \pi_1(Y)=V\in\tau$.
Then $X\cap Y$ is a cofinite extension of $U\cap V\in\tau$,
so $X\cap Y\in\widehat{\tau}$.

(Arbitrary union)
If $X_i\in\widehat{\tau}$, then they are cofinite extensions of
some $U_i\in\tau$.
Then $\cup X_i$ is a cofinite extension of $\cup U_i\in\tau$,
so $\cup X_i\in\widehat{\tau}$.

($T_1$)
Every cofinite subset of $\kappa\times\kappa$
is a cofinite extension of $\kappa$,
so any topology of the form
$\widehat{\tau}$ on $\kappa\times\kappa$
contains all cofinite sets.
This means that any such topology is $T_1$. \\\

**Sketch of proof of II.**
Given topologies $\tau_i$ on $\kappa$ we must argue that

(Inj) the map $\tau\mapsto \widehat{\tau}$ is injective,

(M) $\widehat{\bigcap \tau_i}=\bigcap\widehat{\tau_i}$, and

(J) $\widehat{\bigvee \tau_i}=\bigvee\widehat{\tau_i}$.

The map $\tau\mapsto \widehat{\tau}$ is
is easily seen to be order-preserving (and 1-1), so I focus on the claims

(M)' $\widehat{\bigcap \tau_i}\supseteq\bigcap\widehat{\tau_i}$, and

(J)' $\widehat{\bigvee \tau_i}\subseteq\bigvee\widehat{\tau_i}$.

Let's start with (M)'. Choose a set
$X\in \bigcap\widehat{\tau_i}$ and let $U=\pi_i(X)$.
Then $U\in \bigcap \tau_i$ for all $i$,
and $X$ is a cofinite extension of $U$,
so $X\in \widehat{\bigcap\tau_i}$.

Now (J)'. Suppose that $X\in \widehat{\bigvee\tau_i}$.
Then $X$ is a cofinite extension of some set in $\bigvee\tau_i$,
and a typical such set has the form
$\bigcup_i (U_{i1}\cap \cdots\cap U_{ik_i})$ where $U_{ij}\in\tau_j$.
In will now suffice for us to show that $X$ can also be represented
in the form
$\bigcup_i (\overline{U}_{i1}\cap \cdots\cap \overline{U}_{ik_i})$
where $\overline{U}_{ij}$ is a cofinite
extension of some set in some $\tau_j$.
Of course, we will choose
$\overline{U}_{ij}$ to be a
cofinite extension of the set $U_{ij}\in\tau_j$,
but we must explain how to choose the fibers
of $\overline{U}_{ij}$.
If some $x\in U_{ij}$ also belongs
to $\pi_1(X)$, then choose the fiber
over $x$ in $\overline{U}_{ij}$
so that it agrees with the fiber over $x$ in $X$,
which must be cofinite in $\kappa$.
For any other $x\in U_{ij}$
it doesn't matter how you choose
the fiber over $x$ in $\overline{U}_{ij}$
except that it must be cofinite.
(To be specific, choose this fiber to be all of $\kappa$.)

We have now chosen $\overline{U}_{ij}\in\widehat{\tau_j}$
so that $\bigcup (\overline{U}_{i1}\cap \cdots\cap \overline{U}_{ik_i})$
has the same first projection
and the same fibers as $X$, hence it equals $X$.
This represents $X$ as an element of $\bigvee \widehat{\tau_i}$. \\\