# 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.

• Realize that the image of $f^*$ is the same as the image of $f$ . If $f$ is not surjective (and you don't demand that in your question) then $f(X)≠Y$ . This means that no set $O\in\tau'$ can exist with $f^*(O)=Y$ wich contradicts that $f^*(\tau')$ is a topology. – drhab Jul 30 '15 at 9:59
• @drhab I see, thanks! So do you think it is a mistake in the book? I mean, maybe the author wanted to show some other concept but end up with having a typo in this paragraph, or do you think I should just let it go and ignore this part of the section? – HSea12345n Jul 30 '15 at 13:49
• Was it explicitly said in the book that $f^*(\tau')$ is a topology in this context? I have my doubts about that (and have the book not at my disposal). I suspect this all has to do with writing the continuous $f$ as $f=i\circ s$ with $i$ injective and $s$ surjective (in your question $i=f^*$ and $s=\pi_f$). If $Z$ is domain of $i$ and codomain of $s$ then the question rises: what topologies on $Z$ make $i$ and $s$ both continuous? There is a finest (the quotienttopology=identificationtopology) and a coarsest (in your question the topology $f^{*-1}(\gamma)$). – drhab Jul 30 '15 at 14:34

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:$$\tau_{Z}^{\text{coarse}}\subseteq\tau_{Z}\subseteq\tau_{Z}^{\text{fine}}$$

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: $$\tau_Z=\tau_Z^{\text{coarse}}\iff i\text{ is an embedding}$$
• Thank you so much for the clear and insightful explanation! The book didn't explicitly say that $f^{*}(\tau')$ is a topology, even though it used some notation such as $(Y,f^{*}(\tau'))$. but now after reading this, I'm pretty sure it's a mistake on the book. I think the author's original intention was to show the comparison of different topologies on $X/\sim_{f}$ instead of on $Y$. Anyway, Thanks! – HSea12345n Jul 30 '15 at 16:13