Let's say that $A \subset X$ is a deformation retract. It follows that $A$ is both a retract and a space homotopically equivalent to $X$. Is the converse true? Probably not, but I couldn't find any example yet.

More specifically the converse would be:

If $A \subset X$ is a retract which is homotopic to $X$ as a topological space then does there exist a homotopy between the retraction and the identity map: $$H:X \times [0, 1] \to X$$ such that $H(x,0)=x$, $H(x,1)\in A$ and $H(a,1)=a$ for $a\in A$.

  • $\begingroup$ You should probably specify whether you mean a strong deformation retract or a weak deformation retract. $\endgroup$ – Hew Wolff May 23 '16 at 3:14
  • $\begingroup$ @HewWolff I've updated the question. $\endgroup$ – freakish May 23 '16 at 4:37

No. Let $X = \{0,1,2,3,\dots\}$ and $A = \{1,2,3,\dots\}$, both with the discrete topology, and let $i: A \to X$ be the inclusion. Then $i$ has a retraction $r: X \to A, n\mapsto\max\{n,1\}$, and is even a cofibration. $X$ and $A$ are clearly isomorphic. The inclusion, however, is not a homotopy equivalence.


If your question is whether every space $Y$ which is homotopically equivalent to a space $X$ must be a retract of $X$ then this is certainly not true. One main requirement for a deformation retract is that $A \subset X$. One can easily produce homeomorphic spaces which aren't subspaces of each other. Consider two disjoint disks of different radii in the plane. Does this answer your question?

  • $\begingroup$ A more plausible converse would be "if $A \subset X$ is a retract of $X$ and is homotopically equivalent to $X$, then $A$ is a deformation retract". $\endgroup$ – Hew Wolff May 23 '16 at 2:52
  • $\begingroup$ Sorry, I'm asking about the converse.See the updated question. $\endgroup$ – freakish May 23 '16 at 4:38

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