# If $r : X \to A$ is a deformation retract and $i : A \to X$ is inclusion, then $i(r)$ is homotopic to $id$ on $X$

If $r : X \to A$ is a deformation retract and $i : A \to X$ is inclusion, then $i(r)$ is homotopic to $id$ on $X$.

I know that since r is a deformation retract, it is homotopic to id on X, but I don't see an obvious construction of the above homotopy.

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What is your definition of a deformation retract? Usually, the property you're asking about is contained in the definition, see en.wikipedia.org/wiki/… – t.b. Mar 20 '11 at 2:59
How do you prove the equivalence of those two definitions? – Dan Donnelly Mar 20 '11 at 3:09
Which direction isn't clear? $r = F(\cdot,1)$ and conversely $\operatorname{id} \simeq ir$ gives you a map $F: X \times [0,1] \to X$ such that $F(x,0) = x$ and $F(x,1) = ir(x)$, in particular $F(a,1) = ir(a) = a$ since $r$ is a retraction. – t.b. Mar 20 '11 at 3:24
I'm having trouble seeing that if $\operatorname{id} \simeq r$ and $r$ is a retract, then $\operatorname{id} \simeq ir$. – Dan Donnelly Mar 20 '11 at 3:40
Ah, I think I see the confusion: If you're thinking of $r$ as a function $r: X \to A$ then strictly speaking the statement $\operatorname{id} \simeq r$ doesn't make sense because $\operatorname{id}: X \to X$ and $r: X \to A$ are not maps between the same spaces. The relation $f \simeq g$ only makes sense for $f,g: X \to Y$, that is to say maps between the same spaces. What is tacitly assumed when you write $\operatorname{id} \simeq r$ is that you view $r$ as a function $X \to X$, so actually $r$ is an abuse of notation for $ir$. – t.b. Mar 20 '11 at 3:46

Recall that a retraction $r: X \to A$ is a map such that $ri = \operatorname{id}_{A}$, where $i: A \to X$ is the inclusion.

There are two definitions of a deformation retract:

1. The map $r: X \to A$ is called a deformation retraction if $r$ is a retraction and $ir \simeq \operatorname{id}_{X}$.

2. The subspace $A$ of $X$ is called a deformation retract if there is a function $F: X \times [0,1] \to X$ such that $F(x,0) = x$ and $F(x,1) \in A$ for all $x \in X$ and $F(a,1) = a$ for all $a \in A$. More accurately, we should say that $F(x,1)$ factors as $ir$ for $r: X \to A$ (necessarily unique since $i$ is injective) and write the last condition as $F(i(a),1) = i(a)$ since we're viewing $A$ as a space in its own right.

The two definitions are equivalent.

If $F$ is as in $2$ then we can write $ir = F(\cdot,1)$ and $r$ is a retraction by the properties of $F$. Clearly $F$ is a homotopy between $\operatorname{id}_X$ and $ir$.

Conversely, the hypothesis $\operatorname{id} \simeq ir$ gives a map $F: X \times [0,1] \to X$ such that $F(x,0) = x$ and $F(x,1) = ir(x)$. Since $r$ is a retraction we have $F(i(a),1) = ir(i(a)) = i(a)$.

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