# Every absolute retract (AR) is contractible

This is homework.

I need to show that every AR is contractible.

All I can basically do here is list definitions:

A space $Y$ is AR if: $X$ is metrizable, $A$ is closed subset of $X$ and $f: A \mapsto Y$ is continuous, then $f$ has a continuous extension $g: X \mapsto Y$.

A space $Y$ is contractible if $id_Y :Y \mapsto Y$ is homotopic to a constant map.

I was thinking if I could use the fact that every AR is path connected and also every contractible space is path connected.

#### Solutions Collecting From Web of "Every absolute retract (AR) is contractible"

You need to find a continuous function
$$h\colon Y\times[0,1]\to Y$$
and $y_0\in Y$ such that $h(y,0)=y$ and $h(y,1)=y_0$.
Thus it suggests itself to let $X=Y\times[0,1]$ and $A=Y\times\{0,1\}$.
Of course $A$ is closed in the product topology because $Y$ is closed in $Y$ and $\{0,1\}$ is closed in $[0,1]$.
There “only” remains the question whether $X$ is metrizable. Can you take over from there?

AR are metrizable. In fact AR = AR(M) = M intersect with ES(M) where M = class of metric spaces and ES(M) class of absolute extensors for metric spaces, AR(M) = absolute retracts for metric spaces..