It can be shown that the infinitedimensional rational projective space $\mathbb{QP}^\infty$ is a connected, Hausdorff topological space. What can be said about its homotopy type (is it simply connected, is there any hope to compute its cohomology algebra)?

1$\begingroup$ I would guess that it's totally pathdisconnected, and so homotopically discrete. But that's just a guess. $\endgroup$ – Tim Campion Dec 22 '17 at 20:33

2$\begingroup$ A path from $(a,\dots)$ to $(b,\dots)$ projects to a path from $a$ to $b$ in $\mathbb Q$, which is impossible if $a\neq b$. $\endgroup$ – Fan Zheng Dec 22 '17 at 20:35

1$\begingroup$ @FanZheng I'm not sure about anything, I'm just sitting here waiting for Eric to answer :) $\endgroup$ – Fosco Dec 22 '17 at 20:37

$\begingroup$ @FanZheng maybe you saying that a path $\gamma : I \to \mathbb{QP}^\infty$ lifts along the projection from $\mathbb Q^\infty$? And the domain now is totally disconnected, I agree. $\endgroup$ – Fosco Dec 22 '17 at 20:43

7$\begingroup$ Asking for the weak homotopy type of $\mathbb{QP}^\infty$ is the wrong question to ask, because that topological space does not have enough maps $\mathbb R^n\to \mathbb{QP}^\infty$. Instead, one should ask for its shape: en.wikipedia.org/wiki/Shape_theory_(mathematics) $\endgroup$ – André Henriques Dec 22 '17 at 21:48
Any countable Hausdorff space $Q$ is totally pathdisconnected. Indeed, if $f:[0,1]\to Q$ is continuous, then its image $X$ is a countable connected compact Hausdorff space. By Urysohn's lemma, then, continuous maps from $X$ to $[0,1]$ separate points. But $X$ is connected, so the image of a continuous map from $X$ to $[0,1]$ is connected, and so must be just a single point since $X$ is countable. Thus $X$ can only have one point, so $f$ is constant.
So, in particular, $\mathbb{Q}\mathbb{P}^\infty$ is totally pathdisconnected, and has the weak homotopy type of a countable discrete space.
(In fact, more strongly, any countable $T_1$ space is totally pathdisconnected. See Why are the integers with the cofinite topology not pathconnected?)

1

$\begingroup$ Is it true that the square $$ \begin{array}{ccc} \mathbb Q^\infty &\to &\mathbb R^\infty\\ \downarrow && \downarrow \\ \mathbb{QP}^\infty &\to &\mathbb{RP}^\infty\\ \end{array} $$ commutes? In principle, the quotient by multiplication for a nonzero rational gives a different quotient than multiplication for a nonzero real number; but I think the quotients are the same by density. $\endgroup$ – Fosco Dec 22 '17 at 21:41

1$\begingroup$ That square certainly commutes. Maybe your question is whether the bottom map is an inclusion of a subspace? I believe the answer is yes again, but it takes a bit of work to proveI don't think it follows automatically from density or anything like that. (Note, though, that the space $\mathbb{RP}^\infty$ here is not the space that is normally called $\mathbb{RP}^\infty$: the usual $\mathbb{RP}^\infty$ comes from giving a colimit topology to $\mathbb{R}^\infty$, not the product topology.) $\endgroup$ – Eric Wofsey Dec 22 '17 at 22:31
$QP^\infty$ has an open cover by copies of $Q^\infty$. If $\gamma: I \to QP^\infty$ is a path, this pulls back to an open cover of the interval $I$, so $I$ has a cover by open intervals, each of which is mapped into some $Q^\infty$, and then, as Fan Zheng points out, projects down to $Q$ and thus is constant. So the path is constant. Hence $QP^\infty$ is homotopically discrete.
It is nice that you asked a question about the space $\mathbb Q P^\infty$. I have thought about this space for a long time and came to the conclusion that $\mathbb Q P^\infty$ is the most "regular" space among countable connected Hausdorff spaces. It seems that $\mathbb Q P^\infty$ is a unique space among countable connected Hausdorff spaces that admits a simle topological characterization:
Theorem. A topological space $X$ is homeomorphic to $\mathbb QP^\infty$ if and only if $X$ is countable, Hausdorff, and has a countable base $\mathcal B$ of the topology such that for any $n\ge 2$ and basic open sets $U_1,\dots,U_n\in\mathcal B$ the intersection $\bar U_1\cap\dots\cap \bar U_n$ is connected, nonempty, and has zerodimensional complement $X\setminus (\bar U_1\cap\dots\cap \bar U_n)$.
The proof can be done by a (moreorless) standard backandforth argument.

$\begingroup$ I'd say that the next question is to speculate about the homotopy type of the groupoid of homeomorphisms $X\cong \mathbb{QP}^\infty$? :) $\endgroup$ – Fosco Dec 27 '17 at 17:20

$\begingroup$ @FoscoLoregian Why groupoid? The set of homeomorphisms carries a natural group structure... $\endgroup$ – Taras Banakh Jan 2 '18 at 10:59

$\begingroup$ Sure, but instead of being a topological group I was thinking about the homotopy type of its classifying space. But there is no real point in my comment :) j $\endgroup$ – Fosco Jan 2 '18 at 11:43
It would be more appropriate to study it using algebraic geometry; say, etale homotopy theory. See:
Čech Theory: Its Past, Present and Future http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.rmjm/1250128825&view=body&contenttype=pdf_1
Dave

$\begingroup$ Hi prof. Edwards, thanks for your answer. I expect indeed that etale homotopy helps, here, as much as shape theory. But my knowledge of these topics has empty interior. Feel free to add a few remarks on what can be said. $\endgroup$ – Fosco Dec 22 '17 at 22:22

1$\begingroup$ As it stands, this answer is not very helpful. Do you have something concrete to say about the shape of $\mathbb{QP}^\infty$. E.g., is it contractible? $\endgroup$ – André Henriques Dec 23 '17 at 17:00