# Solid spaces are locally contractible

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?

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

For each open neigborhood $$U$$ of $$y_0$$ in $$Y$$ there exists an open neighborhood $$V$$ of $$y_0$$ in $$Y$$ contained in $$U%$$ such that the inclusion $$V \hookrightarrow U$$ is null-homotopic.

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.