# 4.02 Homotopy extension property (HEP)

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

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# 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 \theta-graph
and the figure 8 are homotopic ``just by squishing the middle bar of
the \theta 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.