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If $A\subset X$ is a deformation retract of $X$. Are $X$ and $A$ homotopy equivalent?

Stefan Hamcke
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Ronald
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2 Answers2

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If yours is the standard definition, the answer is "yes".

Namely, $A \subset X$ is a deformation retract of $X$ if there is a continuous map $r: X \longrightarrow A$ such that $r\circ i = \mathrm{id}_A$, where $i: A \longrightarrow X$ is the inclusion, and $i\circ r \simeq \mathrm{id}_X$.

Since $=$ implies $\simeq$ and being of the same homotopy type means just that there are maps $f: A \longrightarrow X$ and $g: X \longrightarrow A$ such that $f \circ g \simeq \mathrm{id}_X$ and $g\circ f \simeq \mathrm{id}_A$, obviously being a deformation retract implies being homotopy equivalent, or having the same homotopy type.

Agustí Roig
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Let's simply say that there are many different kind of deformation retraction, one stronger than the other.

The weaker form states that $A \subseteq X$ is a (weak)deformation retract of $X$ iff there's a map $r \colon X \to A$ such that $r$ is both a left and right homotopy inverse to the inclusion map $i \colon A \to X$ (so $A$ must be homotopy equivalent to $X$).

The stronger form states that $A$ is a deformation retract of $X$ iff exists a map $D \colon X \times I \to X$ such that $D(a,t)=a$ for every $a \in A$, $D(x,0)=x$ and $D(x,1) \in A$ for all $x \in X$ (i.e. $D$ is an homotopy relative to the subspace $A$ between the identity and a map of the form $i\circ r$ for some $r \colon X \to A$). Since $D$ is relative to $A$ you get that the $r(a)=a$ for all $a \in A$, so $r \circ i = 1_A$, while $D$ is an homotopy between $1_X$ and $i \circ r$, so $r$ and $i$ are homotopy equivalences.

Giorgio Mossa
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    This is a great answer, but can you explain $D$ is relative to $A, r(a)=a$.... I am not sure if there is anything to explain actually but just not completely understanding why $r\circ i = 1_A$. – grayQuant Dec 09 '15 at 00:31
  • @grayQuant $D$ is relative to $A$ means that $D$ keep fixed the points of $A$: by definition $\forall a \in A, t \in I$ we have $D(a,t)=a$. Since $D(x,1)=i\circ r(x)$ we have that for $x=a \in A$ it must be $i \circ r(a) = D(a,1)=a$ and since $i$ is an embedding (that is $i(x)=x$ for every $x \in A$) we have that $r(a)=a$ (this proves that $r$ fixes the points of $A$). Hope this helps. – Giorgio Mossa Dec 10 '15 at 10:38
  • Does the stronger version imply the weaker version? –  Feb 14 '22 at 10:40
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    @user974406 absolutely: basically weak deformation retract requires to have an homotopy between $1_X$ and the composite $i \circ r$, and an homotopy between $1_A$ and $r \circ i$. The strong definition provides you with an homotopy between $1_X$ and $i \circ r$ (one that keeps points of $A$ fixed during deformation) and the trivial homotopy between $1_A$ and $r \circ i$, this homotopy is trivial since in the strong version we have $r \circ i=1_A$. – Giorgio Mossa Feb 20 '22 at 17:29