I have a question while reading Steenrod's The Topology of Fibre Bundles, section 12.
A space $Y$ is called solid if, for any normal space $X$, closed subset $A$ of $X$, and map $f:A\to Y$, there exists a map $f':X\to Y$ such that $f'|_A=f$.
Let $Y$ be solid such that $Y\times I$ is normal. Fix a point $y_0\in Y$. Note that $A:=(Y\times 0)\cup (y_0\times I)\cup (Y\times I)$ is a closed subset of $Y\times I$. Define $f:A\to Y$ by $f(y,0)=y$, $f(y,1)=y_0$ and $f(y_0,t)=y_0$. Then solidity of $Y$ implies that $f$ extends to $f':Y\times I\to Y$. Now $f'$ is a homotopy from $\textrm{id}_Y$ to the constant map $Y\to y_0$. Thus $Y$ is contractible. Since $y_0$ is arbitrary, it also follows that $Y$ is locally contractible.
I can't see why $Y$ is locally contractible. How does this argument shows that each point of $Y$ have arbitrary small locally contractible neighborhoods?
Best Answer
A more common notation for a solid space is "absolute extensor for normal spaces".
Your construction of $f'$ shows that $(Y,y_0)$ is pointed contractible for each $y_0 \in Y$. This immediately implies that
If this property is satisfied, $Y$ is called locally contractible at $y_0$. If $Y$ is locally contractible at all of its points, it is called locally contractible.
This is the standard definition. The requirement that each $y_0 \in Y$ has arbitrarily small (open) contractible neigborhoods is stronger and I doubt that is true for all absolute extensors. You should check Steenrod's definition.
See also ANR is locally contractible .