Some basic questions about the rationalization of a top. space Let $X$ be a topological space and let $X_\mathbb Q$ be its rationalization.
1) What is the rationalization of $X_\mathbb Q$, is it   $X_\mathbb Q$ itself?
2) If $X$ is a CW-complex, does that imply necessarily that  $X_\mathbb Q$ is also a CW-complex?
3) If  $X_\mathbb Q$ and  $X'_\mathbb Q$ are two rationalizations of $X$, how do they relate? My guess is that they are weakly equivalent, and so if they are CW-complexes, then they are homotopy equivalent.
 A: Let $X$ be a simply connected space. Define the rationalization of $X$ as a pair $(X_{\mathbb{Q}},f)$ where $X_{\mathbb{Q}}$ is a rational simply connected space and $f:X\to X_{\mathbb{Q}}$ induces an isomorphism in homotopy $$\pi_*(X)\otimes \mathbb{Q}\to \pi_*(X_{\mathbb{Q}}).$$
There always exists the rationalization (up to homotopy) of such a space $X$. The answers to your questions are:


*

*Let $X$ be a simply connected topological space and consider its rationalization. Using the definition given above, a rationalization of the space $X_{\mathbb{Q}}$ can be the pair $(X_{\mathbb{Q}},1_{X_{\mathbb{Q}}})$, because the identity induces an isomorphism $$1_{\pi_*(X_{\mathbb{Q}})}:\pi_*(X_{\mathbb{Q}}) \cong \pi_*(X_{\mathbb{Q}})\otimes \mathbb{Q} \to \pi_*(X_{\mathbb{Q}}).$$

*Yes, you can achieve this. Let $X$ be a simply connected CW-complex. Then, there exists a relative CW-complex $(X_\mathbb{Q},X)$ in such a way that the inclusion $i:X\hookrightarrow X_\mathbb{Q}$ is a rationalization.
In fact, you can build $X_\mathbb{Q}$ by cellular approximation, by induction in the skeleta $X(n)$ of $X$. For this, you have to work out first the rationalization of the sphere, $\mathbb{S}^n_\mathbb{Q}$, as you will have to work with attaching maps $\mathbb{S}^n\to X(n)$.

*Yes. The rationalization $(X_\mathbb{Q},f)$ of a simply connected space $X$ has the following universal property: given any $g:X\to Y$, where $Y$ is a simply connected rational space, there exists a unique (up to homotopy) map $h:X_\mathbb{Q}\to Y$ making the obvious diagram commute. Using this universal property with both rationalizations settles the question.

