Does the lattice of all topologies embed into the lattice of $T_1$-topologies? Let $\kappa$ be an infinite cardinal, and let $\text{Top}(\kappa)$ be the lattice of all topologies on $\kappa$, ordered by $\subseteq$. Let $\text{Top}^{T_1}(\kappa)$ be the lattice of all $T_1$-topologies on $\kappa$.
Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$?
 A: Claim: Any such $\varphi$ would have to map into a set on which all homomorphisms of $\text{Top}^{T_1}(\kappa)$ are constant.

This follows from Theorems 1 and 2 of

Hartmanis, Juris, On the lattice of topologies, Can. J. Math. 10, 547-553 (1958). ZBL0087.37403.
  (See https://cms.math.ca/openaccess/cjm/v10/cjm1958v10.0547-0553.pdf)

Theorem 1 says that $\text{Top}(\kappa)$ has only trivial homomorphisms (they are either constant or embeddings).
Theorem 2 says that $\text{Top}^{T_1}(\kappa)$ does have some nontrivial ones; so let $\psi$ be one of them.
If the Claim failed then $\psi\circ\varphi$ would be a nontrivial homomorphism of $\text{Top}(\kappa)$, contradicting Theorem 1.
A: This is a long comment about Bjørn's answer rather than an answer itself.
Hartmanis proves
Thm 1 If $\kappa>2$, then $\text{Top}(\kappa)$ is a simple lattice.
Thm 2 If $\kappa$ is infinite, then 
(1) For any finite subset $F\subseteq \kappa$, the restriction map 
$\rho_F:\tau\mapsto \tau|_{\kappa-F}$ is a nonconstant, noninjective, complete lattice homomorphism from 
$\text{Top}^{T_1}(\kappa)$ to $\text{Top}^{T_1}(\kappa-F)\;(\cong\text{Top}^{T_1}(\kappa))$. 
(2) For any finite subset $F\subseteq \kappa$, $\ker(\rho_F)$ is a proper, complete congruence on $\text{Top}^{T_1}(\kappa)$. Every proper, complete congruence has this form. If $F\neq G$ are distinct finite subsets of $\kappa$, then $\ker(\rho_F)\neq\ker(\rho_G)$. \\\
Theorem 1 implies that if $\varphi:\text{Top}(\kappa)\to\text{Top}^{T_1}(\kappa)$ is an embedding, then for any congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ one must have either 
Case 1. the image of $\varphi$ is contained in a single $\theta$-class, or
Case 2. the image of $\varphi$ is contained in a $\theta$-transversal.
Otherwise $\varphi^{-1}(\theta)$ would be a nontrivial proper congruence of a simple lattice. Thus each congruence $\theta$ on $\text{Top}^{T_1}(\kappa)$ restricts the possibilities for $\varphi$.
Theorem 2 implies that there are lots of $\theta$'s to choose from, namely all of the form $\ker(\rho_F)$. 
Thms 1&2 imply that there are lots of restrictions $\varphi$ must satisfy, but there are not enough restrictions to rule out the existence of $\varphi$. In my answer on this page I construct $\varphi$ whose image is contained in a $\theta$-transversal for every $\theta$ of the form $\ker(\rho_F)$. 
A: 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 $\bigvee \widehat{\tau_i}$. \\\
