Corollary 22.3 in Munkres's book In Topology, the second edition by Munkres, in section 22, on page 142 he says the following:
"Theorem 22.2. Let $p:X\rightarrow Y$ be a quotient map. Let $Z$ be the space and let $g:X\rightarrow Z$ be a map that is constant on each set $p^{-1}( \{ y \} )$, for $y\in Y$. Then $g$ induces a map $f:Y\rightarrow Z$ such that $f\circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Corollary 22.3. Let $g:X\rightarrow Z$ be a surjective continuous map. Let $X^*$ be the following collection of subsets of $X$: $$X^*=\left \{ g^{-1}\left ( \left \{ z \right \} \right )\mid z\in Z \right \}.$$ Give $X^*$ the quotient topology.
(a) The map $g$ induces a bijective continuous map $f:X^*\rightarrow Z$, which is a homeomorphism if and only if $g$ is a quotient map.
Proof. By the preceding theorem, $g$ induces a continuous map $f:X^*\rightarrow Z$; it is clear that $f$ is bijective. Suppose that $f$ is a homeomorphism. Then both $f$ and the projection map $p:X\rightarrow X^*$ are quotient maps, so that their composite $g$ is a quotient map. Conversely, suppose that $g$ is a quotient map. Then it follow from the preceding theorem that $f$ is a quotient map. Being bijective, $f$ is thus a homeomorphism."
My question is:

*

*I don't understand the word "constant" in the sentence "$g:X\rightarrow Z$ be a map that is constant on each set $p^{-1}( \{ y \} )$, for $y\in Y$". What does it mean?

*In the proof, why is f bijective? As far as I know, f is just surjective.

Can someone help me? Thanks.
 A: The sentence ” let $g:X\rightarrow Z$ be a map that is constant on each set $p^{-1}( \{ y \} )$, for $y\in Y$" means that for each $y\in Y$ there is some value $c$ in the codomain of $g$ such that for all $ x \in p^{-1}(\{y\})$ (i.e. all $x$ s.t. $p(x)=y$) we have $g(x) =c$. So if the values of $p$ agree for any two points, then so do their $g$ values.
The last part is just realising what $f$ actually is. The injectivity is then obvious.
A: Regarding your question 1, the thing to observe is that the set
$$p^{-1}(\{y\}) = \{x \in X \mid p(x)=y\}
$$
is a subset of $X$. Given any subset $A \subset X$, to say that "$g : X \to Z$ is constant on $A$" means that the restricted function $g \mid A : A \to Z$ is a constant function. So the statement in question says: for every $y \in Y$, the restriction of the function $g : X \to Z$ to the subset $p^{-1}(\{y\}) \subset X$ is a constant function.
Regarding your question 2, to prove $f$ is injective, consider two elements of $X^*$, call them
$$g^{-1}(\{z_1\}), g^{-1}(\{z_2\}) \in X^*
$$
Assume that
$$f(g^{-1}(\{z_1\})) = f(g^{-1}(\{z_2\}))
$$
The map $f$ which is induced by $g$ is defined by the formula
$$f\bigl(g^{-1}(\{z\})\bigr) = z
$$
Therefore, $z_1=z_2$. From this it follows that
$$g^{-1}(\{z_1\}) = g^{-1}(\{z_2\})
$$
which proves that $f$ is injective.
A: *

*$S = p^{-1}(\{y\})$ is some subset of $X$.  For $g$ to be constant on $S$: for any pair of points $s_1, s_2 \in S$, $g(s_1) = g(s_2)$.  That is, for each $y$ in the image of $p$, there exists $z_y \in Z$ such that $\forall s \in p^{-1}(\{y\}), g(s) = z_y$.


*Let $x,y \in X^*$.  That is, $x \subseteq X$ on which $g$ is constant and likewise for $y$.  If $f(x) = f(y)$, then $x = g^{-1}(f(x)) = g^{-1}(f(y)) = y$, so $x = y$.  This shows the induced $f$ is injective.
A: *

*Constant on each set $p^{-1}( \{ y \} )$ means that the restriction $p \mid_{p^{-1}( \{ y \})} : p^{-1}( \{ y \}) \to Z$ is a constant map. In other words, $p(x) = p(x')$ for all $x, x' \in p^{-1}( \{ y \} $.


*The subsets $g^{-1}(\{z\})$ with $z  \in Z$ are pairwise disjoint, i.e. form a partition of $X$.
The map $f : X^* \to Z$ is defined by $f(g^{-1}( \{ z \} )) = z$. Note that in this definition each $g^{-1}( \{ z \}) $ is an element of $X^*$. Recall that $g^{-1}( \{ z \}) \ne g^{-1}( \{ z' \}) $ for $z  \ne z'$, thus there is no ambiguity in the definition of $f$. You know that $f$ is surjective. Now let $f(g^{-1}( \{ z \} )) = f(g^{-1}( \{ z' \} ))$. By definition this means $z = z'$ and therefore $g^{-1}( \{ z \}) = g^{-1}( \{ z' \} )$. This shows that $f$ is injective.
Another way to see that $f$ is a bijection is this: Define $\phi : Z \to X^*, \phi(z) = g^{-1}(\{z\})$. Then $f(\phi(z)) = z$ and $\phi(f( g^{-1}(\{z\}))) =  g^{-1}(\{z\}$. This shows that $f$ and $\phi$ are inverse to each other.
