clarification of an example of bijection which is not an isomorphism I am working on theory of categories.I got maybe a simple question. 
Let $X$ and $Y$ two topological spaces so we could construct a continuous and injective function $f:X\rightarrow Y$ with dense image which is bijective but not an isomorphism. Could anyone explain to me why is this true?
Thanks.
 A: A continuous injective function with dense image is not necessarily surjective, as the embedding, mentioned in the comments, $\Bbb Q\hookrightarrow\Bbb R$ shows.
Neither a bijective continuous function need be a homeomorphism (isomorphism in the category of topological spaces and continuous functions). 
Because, consider two topologies $\tau_1\subsetneq\tau_2$ on the same set $X\ $ (so that a set being open in $\tau_1$ implies it's open in $\tau_2$, but there's at least one $\tau_2$-open set which is not open in $\tau_1$). 
Then, the identity map $(X,\tau_2)\to(X,\tau_1)$ is continuous and bijective, but its inverse (which is the same mapping, the identity, but with domain and codomain exchanged), is not continuous.
(In the comments $\tau_2$ is taken to be the discrete topology, i.e. the whole $\mathcal P(X)$.)
By the way, up to homeomorphism, this is the only possible occurance of a continuous bijection: let $f:X\to Y$ be continuous bijection, so $f^{-1}$ exists as a function, then pulling back the topology $\tau_Y$ of $Y$ to $X$ along $f^{-1}$, by setting $f^{-1}[\tau_Y]:=\{f^{-1}(V):V\in\tau_Y\}$,  we obtain another topology on $X$ besides the given $\tau_X$, satisfying $\ f^{-1}[\tau_Y]\ \subseteq\ \tau_X$.
We get equality iff $f$ is a homeomorphism.
However, of course, we cannot state that no continuous bijections are homeomorphisms. In contrary, homeomorphisms exist and every homeomorphism must be a continuous bijection (with continuous inverse).
Note that this (lack of) property is special for the topological spaces; concrete algebraic categories do satisfy it: a bijective homomorphism is always an isomorphism.
