Here is an answer to the question as literally asked.

For a function $f: X \to Y$, let $\ker(f) = \{(x, x') \in X \times X: f(x) = f(x')\}$. This is of course an equivalence relation.

**Proposition:** For $f, g: X \to Y$, we have $f \leq_t g$ iff $\ker(g) \subseteq \ker(f)$.

**Proof:** For "only if", notice that $g(x) = g(x')$ implies $f(x) = f(x')$ whenever $f$ is of the form $u \circ g$. For "if", define a partial function $u$ from $Y$ to $Y$ by $u(y) = f(x)$ whenever $y$ is of the form $g(x)$; this is well-defined since if $y = g(x)$ and $y = g(x')$, then also $f(x) = f(x')$ by the assumption $\ker(g) \subseteq \ker(f)$. Then extend $u$ to a total function from $Y$ to $Y$ however you please. $\Box$

Hence for $f, g: X \to Y$, we have $f \simeq_t g$ iff $\ker(f) = \ker(g)$.

Letting $\text{Equiv}(X)$ be the set of equivalence relations on $X$, the last observation says that the map $\left(\text{Fct}(X, Y)/\simeq_t\right) \to \text{Equiv}(X)$ that takes the $\simeq_t$ equivalence class $[f]$ to $\ker(f)$ is well-defined and injective.

In the case $Y = X$, it is also surjective by the axiom of choice. In detail, for any $E \in \text{Equiv}(X)$, there exists a section $s$ of the quotient map $r: X \to X/E$ (meaning $r \circ s = 1_{X/E}$), and then $\ker(s \circ r) = E$.

In that case, the map $\ker: \left(\text{Fct}(X, X)/\simeq_t\right) \to \text{Equiv}(X)^{op}$ is an isomorphism by the proposition, and since $\text{Equiv}(X)$ is a complete lattice, so must be $\text{Fct}(X, X)/\simeq_t$.

**Added later:** It seems all we needed in the paragraph above that begins "In the case $Y = X$" is the existence of an injective function $s: X/E \to X$, not that $s$ needs to be a section of $r$. Because just with injectivity, we get $\ker(s \circ r) = \ker(r) = E$. With that in mind, we can say a little more about the situation for general $Y$.

**Proposition:** For general $Y$, the poset $F = \text{Fct}(X, Y)/\simeq_t$ admits infs of nonempty subsets.

Thus the *only* obstruction to $F$ being a complete lattice is that it might not have the inf of the empty set, i.e., a top element $\top$ dominating all others. If $Y$ has cardinality greater than or equal to that of $X$, that $\top$ will exist.

**Proof:** We already saw that $\ker: F \to \text{Equiv}(X)^{op}$ is a poset embedding. Let $\{[X \stackrel{f_i}\to Y]: i \in I\}$ be a nonempty collection inside $F$, and put $E_i = \ker [f_i]$. The inf of the $E_i$ in $\text{Equiv}(X)^{op}$ is the join $E = \bigvee_{i \in I} E_i$ in $\text{Equiv}(X)$, which certainly exists. The cardinality $|X/E|$ is dominated by any $|X/E_i| = |f_i(X)|$, which is dominated by $|Y|$ using the image factorization $f_i = \left(X \to f_i(X) \subseteq Y\right)$. Hence there is an injection $i: X/E \to Y$, and then $\ker f = E$ for $f = \left(X \twoheadrightarrow X/E \stackrel{i}{\to} Y \right)$, making $[f]$ the inf of the $[f_i]$. $\Box$

isa lattice. $\endgroup$ – Wojowu Jan 13 at 16:59