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Observe that any retract of $\newcommand{\RR}{\mathbb{R}} \RR^n$ is necessarily a closed subspace of $\RR^n$. Assuming this necessary condition, the answer to the question is affirmative. More precise …

$\newcommand{\set}[1]{\lbrace #1 \rbrace}$I will assume that the notation $\Sigma X$ in the question denotes the unreduced suspension of the space $X$. Quick answer: The notion of homotopy equivalenc …

$\newcommand{\into}{\hookrightarrow}$It seems that if $Z$ has the indiscrete topology, then the evaluation map $ev_0 : Z^I \to Z$ has the right lifting property with respect to any map. That provides …

Here is a proof which uses only singular homology.$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\To}{\longrightarrow}$$\def\set#1{\lbrace#1\rbrace}$$\newcommand{\Xminusx}{X\ …

$\newcommand{\RR}{\mathbb{R}} \newcommand{\To}{\longrightarrow} \newcommand{\id}{\mathrm{id}}$The example described in Tom Goodwillie's answer to a related mathoverflow question essentially solves thi …

Edit: Zhen Lin incisively observes in a comment below that the category of compact Hausdorff spaces is monadic over the category of sets, hence is cocomplete. That answers the first part of question 1 …

The following statement follows from results of Palais (see theorem 2.3.3 in "On the existence of slices for actions of non-compact Lie groups"). When $G$ is a Lie group, any principal $G$-bundle (in …

I am looking for an example of a well-pointed space in which no (sufficiently small) neighbourhood of the base-point is contractible. As usual, a well-pointed space is a pointed space in which the inc …