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Let $X$ be some top. space and $X/{\raise.17ex\hbox{$\scriptstyle\sim$}} $ a quotient space with quotient map $q$:
\begin{CD}
X @>q>> X/{\raise.17ex\hbox{$\scriptstyle\sim$}}
\end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{\raise.17ex\hbox{$\scriptstyle\sim$}} $.
Is it true that $q$ is then a homotopy equivalence?
Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?
Best Answer
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $\mathbb{N}$. Given any partition of $\mathbb{N}$ into finite subsets $\mathbb{N}=\coprod_{n \in \mathbb{N}}F_n$, there is the quotient $X/\sim$ where $\sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x \sim y$ iff $x, y \in F_n$for some $n$. Then $X$ and $X/\sim$ are homeomorphic (for example by $n \mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=\mathbb{C}\setminus \mathbb{Z}$, and the quotient $X/\sim$ obtained by shrinking $\{z \in \mathbb{C}\;|\; 0\neq |z| \leq\frac{1}{2}\}$ into a point. Then again, $X/\sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.