# If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $B$ is a deforemation retract of X

If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $B$ is a deformation retract of $X$.

I am a beginner in Algebraic Topology so I tried to write every proof out myself before consulting. I know there are various ways of handling this but I just want to confirm if what I wrote below makes sense.

$A\subset X$ is a deformation retract of X, let $r:X \to A$ such that $r$ is homotopic to the identity map on $X$. Define $H:X\times I \to X$ such that $$H(x,0)=x$$ $$H(x,1)\in A$$ $$H(a,t)=a, a\in A$$ Thus $H(x,0)=x$ and $H(x,1)=r(x)$

Similarly, We have $B\subset A$, a deformation retract of A, Let $s:A \to B$ such that $s$ is homotopic to the identity map on $A$. Define $F:A\times I \to A$ such that $$F(x,0)=x$$ $$F(x,1)\in B$$ $$F(b,t)=B, b\in B$$ Thus $F(x,0)=x$ and $F(x,1)=s(x)$

Now to show that $B$ is a deformation retract of $X$, I define $q=s\circ r :X \to B$.

so $q$ is continuous being the composition of two continuous functions.

Claim: $G:X \times I \to X$. define by $$G(x,t)=F(H(x,t),t)$$ Is the required homotopy between $q$ and the identity on $X$. $$G(x,0)=F(H(X,0),0)=F(x,0)=x$$ $$G(x,1)=F(H(x,1),1)=F(r(x),1)=s(r(x))=s\circ r$$ Is this a good way to go? Any help will be appreciated. thank you.

• This is a good way to go. Jul 3 '18 at 9:33
• @Tyrone Thank you. Jul 4 '18 at 1:34

Your proof isn't quite correct, as there is a subtle domain error: in checking that $$G(x,0) = F(H(x,0),0) = F(x,0) = x,$$ you (implicitly) assume that $$F(x,0) = x$$ for all $$x$$ in $$X$$. However, $$F$$ is defined on $$A \times I$$, while $$x$$ can be anywhere in $$X \supset A$$.
Instead, you can define a deformation retract $$G: X \times I \rightarrow X$$ as follows: \begin{align*} G(x,t) = &H(x,2t), \quad & 0 \leq t \leq \frac{1}{2} \\ G(x,t) = &F(H(x,1),2t-1), \quad & \frac{1}{2} \leq t \leq 1. \end{align*} We will check that $$G$$ properly satisfies the conditions for a deformation retract. First, we have that $$G(x,0)$$ is the identity on $$x$$, $$G(x,0) = H(x,0) = x \;\text{for all}\;x \in X,$$ and that the image of $$G(x,1)$$ is in $$B$$, $$G(x,1) = F(H(x,1),1) = F(a,1) \in B \;\text{for all}\;x \in X, \;\text{some}\; a \in A.$$ Moreover, \begin{align*} G(b,t) = &H(b,2t) = b, \quad & 0 \leq t \leq \frac{1}{2} \\ G(b,t) = &F(H(b,1),2t-1) = F(b,2t-1) = b, \quad & \frac{1}{2} \leq t \leq 1 \end{align*} shows that $$G(b,t) = b$$ for all $$b \in B$$ and hence $$G(x,t)$$ is the identity on $$B$$. All that remains is continuity; in particular, the only potential discontinuity would be at $$t = \frac{1}{2}$$. Computing $$F(H(x,1),2\Big{(}\frac{1}{2}\Big{)} - 1) = H(x,1)$$ shows that the piecewise components of $$G$$ agree at $$t = \frac{1}{2}$$ and hence $$G$$ is continuous.