Questions about the Identification Topology and Equivalence Class from "Introduction to Topology" by Mendelson I am currently reading Introduction to Topology by Bert Mendelson, and I have some questions regarding the topic on Identification Topology in his book.
Let $(X,\tau)$ and $(Y,\gamma)$ be topological spaces.
Let $f:X\rightarrow Y$ be a continuous function.
Define $\sim_{f}$ to be the relation on $X$ by $x\sim_{f}x'$ if $f(x)=f(x')$ and $X/\sim_{f}$ be the set of equivalence classes.
Let $\pi_{f}:X\rightarrow X/\sim_{f}$ be the map that maps each $x\in X$ onto its equivalence class. And also define $\tau'$ to be the identification topology on $X/\sim_{f}$ determined by $\pi_{f}$.
It is true that for $x,x'\in X:$ $\pi_{f}(x)=\pi_{f}(x')$ iff $f(x)=f(x')$. Thus by the theorem, $f$ induces a continuous map $f^{*}:X/\sim_{f} \rightarrow Y$ such that $f=f^{*}\circ \pi_{f}$ by defining $f^{*}(u)=f(x)$ for $u=\pi_{f}(x)$. And one can also show $f^{*}$ is injective.
Define $f^{*}(\tau')=\{f^{*}(O)\subset Y : O\in\tau'\}$
Now my questions are: is it true that $\gamma\subset f^{*}(\tau')$? Will $f^{*}(\tau')$ form a topology on $Y$? If it would, will $f^{*}(\tau')$ be the finest or weakest topology on $Y$ such that $f$ is continuous?
In the original text, it says "Since $f^{*}$ is continuous, $f^{*-1}(\gamma)\subset \tau'$, or equivalently, since $f^{*}$ is one-one, $\gamma \subset f^{*}(\tau')$."..."Thus the topology $\tau'$ carried over to $Y$ by $f^*$ is the weakest or smallest topology such that f is continuous."(From chapter 3, Section 3: Identification Topologies, Page 103, last paragraph)
The implication that $f^{*}$ injective $\implies\gamma \subset f^{*}(\tau')$ was not obvious to me. And also, if that's the case, why would  " $\tau'$ carried over to $Y$ by $f^*$ " (I'm assuming it's referring to $f^{*}(\tau')$) be the smallest topology? 
Thanks.
 A: I hope that this will make things more clear to you.
In this answer $i$ corresponds with the $f^*$ in your question and $s$ with the $\pi_f$ in your question.
Let $X$ and $Y$ be topological spaces and let $f:X\rightarrow Y$
be continuous. Looking at $f$ purely as a function we can write it
as $f=i\circ s$ where $i$ is injective and $s$ is surjective. As
codomain of $s$ and domain of $i$ we can take the set $Z$ that
is denoted in your question as $X/\sim_{f}$.
Now we want $Z$ to be equipped with a topology such that $i$ and
$s$ are both continuous. 
Two extreme candidates are:
1) $\tau_{Z}^{\text{fine}}:=\left\{ O\in\wp\left(Z\right)\mid s^{-1}\left(O\right)\in\tau_{X}\right\} $.
This under the motto: "take every subset of $Z$ in the topology as long as you do
not violate the continuity of $s$".
2) $\tau_{Z}^{\text{coarse}}:=\left\{ i^{-1}\left(U\right)\mid U\in\tau_{Y}\right\} $
or shortly $i^{-1}\left(\tau_{Y}\right)$. 
This under the motto: "leave
every subset of $Z$ out as long as you do not violate the continuity
of $i$."
If $U\in\tau_{Y}$ then $s^{-1}\left(i^{-1}\left(U\right)\right)=f^{-1}\left(U\right)\in\tau_{X}$
since $f$ is continuous, showing that $i^{-1}\left(U\right)\in\tau_{Z}^{\text{fine}}$.
So we have $\tau_{Z}^{\text{coarse}}\subseteq\tau_{Z}^{\text{fine}}$.
The maps $i$ and $s$ are both continuous if and only the topology
$\tau_{Z}$ on $Z$ satisfies:$$\tau_{Z}^{\text{coarse}}\subseteq\tau_{Z}\subseteq\tau_{Z}^{\text{fine}}$$

addendum
This actually gives two different factorizations of the continuous function $f=i\circ s$:
$$\tau_Z=\tau_Z^{\text{fine}}\iff s\text{ is a quotientmap}$$ 
and:
$$\tau_Z=\tau_Z^{\text{coarse}}\iff i\text{ is an embedding}$$
