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Let $κ$ be an infinite cardinal, $S$ a set of cardinality $κ$, and let $I = [0, 1]$ be the closed unit interval. Define an equivalence relation $E$ on $I × S$ by $(x,α) E (y,β)$ if either $x = 0 = y$ or $(x,α) = (y,β)$. Let $H(κ)$ be the set of all equivalence classes of $E$; in other words, $H(κ)$ is the quotient set obtained from $I ×S$ by collapsing the subset ${0}×S$ to a point. For each $x ∈ I$ and each $α ∈ S$, $(x,α)$ denotes the element of $H(κ)$ corresponding to $(x,α) ∈ I × S$. The topology induced from the metric $d$ on $H(κ)$ defined by $d((x,α),(y,β))=|x − y|$ if $α = β$, and $d((x,α),(y,β))= x + y$ if $α\not=β$. The set $H(κ)$ with this topology is called the hedgehog of spininess $κ$ and is often denoted by $J(κ)$. The space is a complete, non-compact, metric space of weight $κ$.

A topological space $X$ is an absolute retract for metrizable spaces $(M)$ provided that it is in $(M)$ and is a retract of each space $Y$ in $(M)$ of which it is embedded as a closed subset.

${\bf Question.}$ Is the hedgehog $J(κ)$ of spininess $κ$ an absolute retract ?

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    $\begingroup$ Yes. It suffices for it to be an absolute extensor. In fact it's an absolute extensor for collectionwise normal spaces (and even characterises this property). This is an exercise in Engelking (see pg.337). $\endgroup$
    – Tyrone
    Mar 27, 2022 at 13:29
  • $\begingroup$ Thanks! It really is. $\endgroup$ Mar 27, 2022 at 14:42

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The answer is positive because the hedgehog $J(κ)$ is $AE$.

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