$X,Y$ be finite topological spaces such that there exist continuous injections from $X$ to $Y$ and $Y$ to $X$ ; are $X$ and $Y$ homeomorphic? Does the analogue of Schroder-Bernstein hold for finite topological spaces ? i.e. Let $X,Y$ be finite topological spaces such that there exist continuous injections from $X$ to $Y$ and $Y$ to $X$ , then are $X$ and $Y$ homeomorphic ? I know that it isn't true for arbitrary topological spaces ( even metric spaces , take $(0,1)$ and $[-1,1]$ ) ; but I don't know what happens if we restrict only to finite topological spaces . Please help . Thanks in advance  
 A: You have a continuous injection from $f:X\hookrightarrow Y$ and a continuous injection $g:Y\hookrightarrow X$.
Since $f$ is injective $|X|\le |Y|$ and since $g$ is, $|Y|\le |X|$. So $|X|=|Y|$ and both are finite. We therefore can conclude that both $f$ and $g$ are bijections.
Let $h = g\circ f : X\to X$. It's a continuous bijection. Since $X$ is finite, the groupe of permutations of $X$ is finite and so $h$ is of finite order, i. e., there is $n\in \Bbb N^*$ so that $h^n=Id_X$. But $(g\circ f)^n=g\circ (f\circ (g\circ f)^{n-1})$. Let $h'=f\circ (g\circ f)^{n-1}$. $h'$ is continuous because $f$ and $g$ are and $g\circ h'=id$ and $g$ is a bijection , so $h'=g^{-1}$ . So $g$ is a bicontinuous bijection and $X$ and $Y$ are homeomorphic.
A: If $X$ and $Y$ are finite, then the assumed continuous injections $X\to Y$ and $Y\to X$ are in fact bijections. Therefore, given a continuous bijection $f:X\to Y$, its inverse maps open subsets of $Y$ injectively to open subsets of $X$, so $X$ has at least as many open subsets as $Y$ does. The other way around of course also holds, so $X$ and $Y$ have equally many open subsets. But that means that the preimages of open subsets of $Y$ are exactly the open subsets of $X$, so our $f$ is in fact a homeomorphism.
A: This is true: 
Suppose that $f:X \longrightarrow Y$ and $g:Y \longrightarrow X$ are continous injections (monomorphisms), where both $X$ and $Y$ are finite topological spaces. 
Then $g \circ f:X \longrightarrow X$ is a injection, and since $X$ is finite, $g \circ f$ is a bijection. Therefore, there exists $n \in \mathbb{N}$ such that $(g\circ f)^n = 1_X$ - the identity on $X$ ($n$ is guaranteed by $X$ being finite, since $g \circ f$ is a permutation on a finite set) . Let $g^* = (g \circ f)^{n-1} \circ g$, then $g^* \circ f = 1_X$. 
Now, by a similar argument, there is $m \in \mathbb{N}$ such that $(f \circ g^*)^m = 1_Y$. Thus we can define  $f^* = (g^* \circ f)^{m-1} \circ f$, then we have $f^*g^* = 1_Y$. Moreover,
$g^*f^* = g^* \circ (f \circ  g^*)^{m-1} \circ f = (g^* \circ f)^m = 1_X$
This shows that $f^*:X \longrightarrow Y$ is a homeomorphic with inverse $g^*$.
