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..