• More explicitly, a set $U \subset X/\sim$ (which is a collection of equivalence classes) is open iff $q^{-1}(U)$ is open in $X$.
• Even more explicitly, $V \subseteq X/\sim$ is open iff $U_V \equiv \{ x \in U : [x] \in V \}$is open in $X$.
• Even more explicitly, we can write $U \equiv \cup_{v \in V} v$, because the elements of $v$ are equivalence classes.

§ Claim: quotient topology is a topology

• The preimage of the empty set is the empty set, and thus is open.
• The preimage of all equivalence classes is the full space, and thus open.
• Preimage of union is union of preimages: $\cup_i q^{-1}(V_i)$ extend $h$ to get a new homotopy $H$: $H_0 = id_X$ and $H_t|A = h_t$.

§ $(X, A)$ have HEP and $A$ is contractible, then $X \simeq X/A$

• Pick $q: X \rightarrow X/A$. We need another map such that their compositions are homotopic to the identities of $X$ and $X/A$.
• Define $s: X/A \rightarrow X$ as a section of $q$, given by $s([a]) \equiv a, s([x]) \equiv x$. This is a section of $q$ since $q \circ s = id_{X/A}$ (That is, $s$ maps entirely within the fibers of $q$).
• Consider $s_t : H_t \circ s : X/A \rightarrow X$. That is, lift from $X/A$ to $X$ using $s$ and then perform $H_t$ on $X$. We claim that The map $(H_1 \circ s)$ is the homotopy inverse of $q$.
• (1a) $(H_1 \circ s) \circ q : X \rightarrow X$ is equal to $H_1$, as $H_1(s(q(A))) = H_1(s([a])) = H_1(a) = a = H_1(A)$, and $H_1(s(q(x))) = H_1(s([x])) = H_1(x)$.
• (1b) So we have $(H_1 \circ s) \circ q = H_1 \simeq H_0 = id_X$, as $H_0 = id_X$ is from defn, and $H_1 \simeq H_0$ is from homotopy. So we are done in this direction.
• (2a) Consider $q \circ (H_1 \circ s) : X \rightarrow X/A$. We wish to show that this is continuous. Let's show that it lifs to a continous map upstairs. So consider $q \circ (H_t \circ s) \circ q : X \rightarrow X/A$. We claim that this is equal to $q \circ H_t$, which is continuous as it is a composition of continuous maps.
• This relationship is hopefully intuitive: $q \circ (H_t \circ s) \circ q$ asks us to treat all of $A$ as if it were $a$ before applying $H_t$. Since $q$ kills whatever $H_t$ does after, and $H_t$ guarantees to keep $A$ within $A$, it's fine if we treat all of $A$ as just $a$. $q \circ H_t$ asks us to treat $A$ as $A$ itself, and not $a$. Since $q$ kills stuff anyway, we don't really care. The real crux of the argument is that $q \circ stab_A = q \circ stab_A \circ s \circ q$ where $stab_A$ is a map that stabilizes $A$.
• (2b) Consider $(q \circ (H_t \circ s) \circ q)(A) = (q \circ H_t \circ s)([a]) = (q \circ H_t)(a)$ --- Since $H_t(a) = h_t(a) = a' \in A$, we crush all data regardless of what happens. This is the same as the value $(q \circ H_t)(A) = [a]$ as $H_t(A) \subseteq A$ and $q(A) = [a]$. For the other set, we get $(q \circ (H_t \circ s) \circ q)(x) = q \circ H_t \circ s([x]) = q \circ H_t(x)$ and hence we are done.
• (2c) Now since $q \circ (H_t \circ s))$ is continuous, and that $q \circ (H_0 \circ s) : X/A \rightarrow X/A = id_{X/A}$, we are done since we can homotope from $q \circ H_1 \circ s \simeq q \circ H_0 \circ s = id_{X/A}$.

Slogan: Use HEP to find homotopy $H$. Use $H_1 \circ s$ as inverse to quotient.