# 4.02 Homotopy extension property (HEP)

## Video

Below the video you will find accompanying notes and some pre-class questions.

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**4.01 CW complexes**. - Next video:
**4.03 CW complexes and the HEP**. - Index of all lectures.

## Notes

*(0.00)* In this section, we will introduce the homotopy extension
property, a useful tool for proving that different spaces are homotopy
equivalent. For example, you can use it to prove that the θ-graph
and the figure 8 are homotopic ``just by squishing the middle bar of
the θ to a point'' (see image below), or the a sphere with a
1-cell attached (joining the North and South poles) is homotopy
equivalent to a pinched torus ``just by squishing the 1-cell to the
pinch point''.

### Homotopy extension property

*(0.58)* Given a space \(X\) and a subspace \(A\), we say that the
pair \((X,A)\) has the homotopy extension property (HEP) if, for
every continuous map \(F\colon X\to Y\) and every homotopy \(h\colon
A\times[0,1]\to Y\) with \(h(x,0)=F(x)\) there exists a homotopy
\(H\colon X\times[0,1]\to Y\) such that \(H(x,0)=F(x)\) and
\(H|_{A\times[0,1]}=h\).

*(2.44)* More colloquially, homotopies of functions defined on \(A\)
extend to homotopies of functions defined on \(X\) (for given initial
data).

*(3.29)* If \((X,A)\) has the HEP and \(A\) is contractible, then
\(X\simeq X/A\), where \(X/A\) denotes the quotient space in which
\(A\) is crushed to a single point.

*(4.40)* To prove that \(X\simeq X/A\), we need to find continuous
maps \(q\colon X\to X/A\) and \(g\colon X/A\to X\) such that
\(g\circ q\simeq id_X\) and \(q\circ g\simeq id_{X/A}\). The map
\(q\colon X\to X/A\) will be the quotient map.

*(5.20)* **Constructing the map \(g\colon X/A\to X\).** The subspace
\(A\) is contractible, so there exists a point \(a\in A\) and a
homotopy \(h_t\colon A\to A\) such that \(h_0=id_A\) and
\(h_1(x)=a\) for all \(x\in A\). Since \(A\subset X\), we can think
of this as a homotopy \(h_t\colon A\to X\).

*(6.14)* Using the HEP for the pair \((X,A)\), we get a homotopy
\(H_t\colon X\to X\) such that \(H_0=id_X\) and
\((H_t)|_{A}=h_t\). At \(t=1\), \(h_1(A)=\{a\}\) and
\((H_1)|_A=h_1\), so \(H_1(A)=\{a\}\). Therefore \(H_1=g\circ q\)
for some continuous map \(g\colon X/A\to X\) (in other words,
\(H_1\) descends to the quotient).

*(7.57)* The fact that \(g\) is continuous follows from this section
on continuous maps out of a quotient space (continuous maps from
\(X/\sim\to Y\) are just continuous maps \(X\to Y\) which descend to
the quotient). This map \(g\) is going to be our homotopy inverse
for \(q\).

*(8.20)* **The map \(g\) is a homotopy inverse for \(q\).** We need to
prove:

*(9.00)*\(g\circ q\simeq id_X\): This holds because \(g\circ q=H_1\simeq H_0=id_X\).*(9.22)*\(q\circ g\simeq id_{X/A}\): Since \((H_t)|_{A}=h_t\), the homotopy \(q\circ H_t\colon X\to X/A\) has the property that \((q\circ H_t)(A)\subset q(h_t(A))\subset q(A)\). Since \(q(A)\) is a single point, this implies that \(q\circ H_t\) factors as \(\bar{H}_t\circ q\) for some continuous map \(\bar{H}_t\colon X/A\to X/A\) (again, using our results on continuous maps out of a quotient space).*(11.22)*We have \(\bar{H}_0=id_{X/A}\) since \(H_0=id_X\). We want to show that \(\bar{H}_1=q\circ g\). We first notice that \[\bar{H}_1\circ q=q\circ H_1=q\circ (g\circ q)=(q\circ g)\circ q,\] which implies \(\bar{H}_1=q\circ g\) if we can cancel the extra \(q\)s on each side of the equation.*(12.34)*We can cancel the \(q\)s because \(q\) is surjective (it's a quotient map) and surjective maps have right-inverses (so can be cancelled from the right).

*(14.05)* We will see in the next video that if \(X\) is a CW complex
and \(A\) is a closed subcomplex then \((X,A)\) has the HEP.

## Pre-class questions

- Assuming that you can use the HEP with impunity, which of the following spaces are homotopy equivalent to one another?

## Navigation

- Previous video:
**4.01 CW complexes**. - Next video:
**4.03 CW complexes and the HEP**. - Index of all lectures.