Let $(M, \tau)$ be a topological space with the following property:
(B) Every bijective map $f: M\to M$ is a homeomorphism.
Under what extra assumption(s) can we conclude that $\tau $ is discrete?
There are several trivial cases that I wish to exclude:
(1) $\tau = \{ \emptyset, M\}$ is trivial: obviously (B) is satisfied:
(2) $M$ is finite: in this case it is easy to show that (assuming that (1) is excluded) $\tau$ is discrete: take a non-empty subset $A$ in $\tau$ with the smallest cardinality. Then (B) implies that all subsets in $M$ with a same cardinality as $A$ is open. Taking finite intersection, we conclude that all singletons is in $\tau$.
Some remarks:
Arguing as in (2), to show that $\tau$ is discrete, it suffices to find one non-empty finite set $A$ in $\tau$.
$T_1$: that is, all singletons in $M$ is closed, is not sufficient: think of the cofinite topology on $M$ (and assume $M$ is infinite), which satisfies (B).
On the other hand, if $(M, \tau)$ is metrizable, then it is discrete (see here). Indeed, we can generalize the proof there and show that $\tau$ is discrete if it is Hausdorff and first countable (a proof is given at the end of the post)
So my (refined) question is:
If $(M, \tau)$ is infinite Hausdorff space and satisfies (B). Is $(M, \tau)$ discrete?
Or, more generally,
Without assuming Hausdorff-ness, What are the assumptions on $(M, \tau)$ to conclude that it is discrete?
We sketch a proof of the following
Theorem: If $(M, \tau)$ is Hausdorff, first countable and satisfies (B). Then $(M, \tau)$ is discrete.
Proof: Assume not, then there is one $a\in M$ so that $\{a\}$ is not in $\tau$. By (B), $\{b\}$ is not in $\tau$ for all $b\in M$. Let $\{U_n\}$ be a local basis at $a$. Using that $(M, \tau)$ is Hausdorff, one can show $\cap _n U_n = \{a\}$. Picking a subsequence if necessary (since $U_n\neq \{a\}$ for all $n$), we assume that $U_{n+1}\setminus U_n$ is non-empty for all $n\in \mathbb N$.
Pick $a_n \in U_n \setminus U_{n+1}$. Similarly we find a local basis $\{V_n\}$ for $b$ and $b_n \in V_n\setminus V_{n+1}$. Note that we can choose $U_1\cap V_1 = \emptyset$ since $(M, \tau)$ is Hausdorff.
Let $\Phi : M \to M$ be the bijective map so that $\Phi (a_n) = b_n$, $\Phi (b_n) = a_n$ for all $n\in \mathbb N$ and $\Phi (y) = y$ otherwise.
By (B), $$\Phi^{-1} (U_1) = (U_1\setminus \{a_n\}) \cup \{b_n\}$$ is open and contains $a$. But $U_n$ is not in $\Phi^{-1} (U_1)$ for all $n$, which is a contradiction. Thus $\{a\}$ is open for all $a\in M$ and hence $(M, \tau)$ is discrete.
Let $\kappa$ be a cardinal which is either infinite or $1$. Then for any set $M$, we can put a topology on $M$ where a set is closed iff either it is all of $M$ or it has cardinality less than $\kappa$. (For $\kappa=1$ this is the indiscrete topology, for $\kappa=\aleph_0$ it is the cofinite topology, for $\kappa=\aleph_1$ it is the cocountable topology, and for $\kappa>|M|$ it is the discrete topology.) Clearly for such a topology, every bijection $M\to M$ is a homeomorphism.
Conversely, I claim that any topology on $M$ for which every bijection $M\to M$ is a homeomorphism is of this form. (So, this means that an axiom implies $M$ must be discrete iff it rules out the possibility that $\kappa\leq |M|$. In particular, for instance, it suffices to assume $M$ is Hausdorff, or even just that there exist two disjoint nonempty open subsets of $M$.) First, note that either every two points are topologically indistinguishable, or every two points are topologically distinguishable by an open set contain one but not the other, for either ordering of the two points. That is, either $M$ is indiscrete or $T_1$. This handles the case where $M$ is finite, so from now on let us assume $M$ is infinite and $T_1$.
Let $\kappa$ be the least cardinal such that $M$ has no proper closed subset of cardinality $\geq\kappa$. First suppose $\kappa\leq|M|$ and suppose $A\subset M$ has $|A|<\kappa$. By definition of $\kappa$, there exists a closed subset $B\subset M$ such that $|A|\leq |B|<\kappa\leq|M|$. We can then find a bijection $f:M\to M$ such that $|B\cap f(B)|=|A|$ (have $f$ fix a subset of $B$ of cardinality $|A|$ and map the rest of $B$ into $M\setminus B$). We can then find a bijection $g:M\to M$ such that $g(B\cap f(B))=A$, and conclude that $A$ is closed. Thus the proper closed subsets of $M$ are exactly those of cardinality less than $\kappa$, as desired.
It remains to handle the case where $\kappa>|M|$, so there is a proper closed subset $A\subset M$ with $|A|=|M|$. Fix a point $x\in M\setminus A$. Since $|A|=|M|$, we can find a bijection $f:M\to M$ such that $f(A)\supseteq M\setminus (A\cup \{x\})$. Then $A\cup f(A)=M\setminus\{x\}$ is closed, so $x$ is isolated. Since $M$ is homogeneous it follows that every point is isolated and $M$ is discrete.