# 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.

-
Every contractible space is path-connected, but very few path-connected spaces are contractible. – Shaun Ault Feb 18 '13 at 17:42
Hint: Let X be the $Y \times I$ but collapse $Y \times \{1\}$ to a point. – Dylan Wilson Feb 18 '13 at 17:44
Are you sure this is the exact definition you are given for AR? Don't you imbed X into Y, from memory, and X is the a deformation retract of Y, or something like that? – gnometorule Feb 18 '13 at 17:46
@gnometorule Yes, that is given as an "alternative" definition in my book. – Valtteri Feb 18 '13 at 17:46
an AR is a metric space by definition. The hint given by Hagen von Eitzen is the correct way to proceed. – user77306 May 11 '13 at 9:08

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?
Well, $Y$ was AR and closed interval is AR also and the countable product of AR is AR, so X is AR... But I doubt that AR implies metrizable. I have a theorem that if $X$ can be embedded to a Hilbert's cube, $X$ is metrizable. I must think this some more. – Valtteri Feb 18 '13 at 18:18
Nope, I cannot really see way to prove $X$ would be metrizable. All the theorems I have seem to demand $X$ to be second countable and regular or so and I can't figure why this would be. If we, however, assume $X$ is metrizable, I think if we define $h(y,0)=y$ and $h(y,1)=y_0$, this is continuous as $h(y,0)$ is identity and $h(y,1)$ is constant, so the preimages stay closed. And it would have the continuous extension, which gives the required properties. – Valtteri Feb 18 '13 at 19:09