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Define $i_X:X\rightarrow X\coprod Y$ as the inclusion mapping $i_X(x)=x$, and similarily $i_Y:Y\rightarrow X\coprod Y$ as the inclusion mapping $i_Y(y)=y$.

Let $X$ and $Y$ be topological spaces. Define a closed subspace $D\subset Y$ with it's inclusion mapping $g:D\rightarrow Y$, and continuous mapping $f:D\rightarrow X$. Let $\sim$ be the equivalence relation on $X\coprod Y$ created by the relation $i_X\circ f(d)\sim i_Y\circ g(d)$ for each $d\in D$.

Show that the equivalence classes in $X\coprod Y$ are of the following:

$\{i_X(x)\}$ if $x\in (X-image(f))$

$\{i_X(f(d))\}\cup i_Y\circ g(f^{-1}(f(d)))$, for $d\in D$

$\{i_Y(y)\}$ if $y\in (Y-D)$

I honestly have no clue how to even attempt this problem. I know that an equivalence class is a set of the form $[a]=\{x\in X:a\sim x\}$ for $a\in X$, but where/how do I start? I apologize for the lack of work and any hints would be much appreciated!

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  • $\begingroup$ What map is $i_Y$? $\endgroup$ – Dog_69 Oct 15 '18 at 22:18
  • $\begingroup$ My apologies, it's just the inclusion mapping associated to $Y$. I edited the post. $\endgroup$ – The math god Oct 15 '18 at 22:19
  • $\begingroup$ When you say the inclusions... do you mean the inclusions into the disjoint union? Because otherwise it is meaningless. I mean $X$ is a subset of $Y$ and $Y$ is a subset of $X$? $\endgroup$ – Dog_69 Oct 15 '18 at 22:22
  • $\begingroup$ Actually, I am pretty sure you're right. They should be inclusions into the disjoint union. My apologies. $\endgroup$ – The math god Oct 15 '18 at 22:26
  • $\begingroup$ Don't worry, it's okay. Now, you should be able to solve cases 1 and 3. $\endgroup$ – Dog_69 Oct 15 '18 at 22:30
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Taking some quotient of a disjoint union is a typical construction in topology, it is the formalization for gluing two spaces $X$ and $Y$ along certain given subspaces/maps.
Maybe an easier example of gluing can help to get the feeling:
Try to formalize the case of gluing two segments together into a circle at their endpoints.

Now, we have $X\sqcup Y$ which consists of points of $X$ (these are $i_X(x)$ for $x\in X$) and points of $Y$ (these are $i_Y(y)$ for $y\in Y$), which are so far separated from each other.
Then, we pick $d\in D\,\subseteq Y$ and glue it to [identify it with] $\ f(d)\,\in X\ $ in the quotient space $X\sqcup Y/\sim$.

(Taking the quotient by an equivalence relation basically forces that relation to actually be equality in the quotient set.)

So, to the questions:

i) If $x$ is not in the image of $f$, then $x$ is not glued to nobody on the other side, so the copy $i_X(x)$ of $x$ will be alone in its equivalence class. (This is by definition of $\sim$.)
iii) The same holds for any $y\notin D$.
ii) Finally, if $d\in D$, then it is now glued to $f(d)$ on the other side, i.e. $i_Y(d)\sim i_X(f(d))$.
Who else are they glued, too?
Note that $\sim$ must be an equivalence relation, and $f$ need not be e.g. injective. Well, if $f(d')=f(d)$ happens for another $d'\in D$, then we have $i_Y(d')\sim i_X(f(d))$ as well, so $i_Y(d')$ is in the same class.
In fact, we collect all those: that is, we take the inverse image $f^{-1}(\, f(d)\,)$.

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  • $\begingroup$ Thank you for your answer. I just have a couple concerns. For i), I am not too sure I understand why the copy of $i_X(x)$ of $x$ will be alone in its equivalence class. Also for ii), I understand the first part, but could you elaborate on the injectivity part? Why take the inverse, and how does that imply $i_Y\circ g(f^{-1}(f(a)))$ is an equivalence class? Thanks again! $\endgroup$ – The math god Oct 15 '18 at 23:23
  • $\begingroup$ Those $i_X(x)$ and $i_Y(y)$ don't appear in the definition of $\sim$ (i.e. after taking symmetric and transitive close of the given relation). The inverse image $f^{-1}( f(d))$ is the set of all $d'$'s which make $f(d')=f(d)$. These $d'$"s are the ones in relation with $d$. (The $g$'s should not bother here, it's simply the inclusion of $D$.) $\endgroup$ – Berci Oct 15 '18 at 23:42
  • $\begingroup$ This actually make so much more sense now. I didn't really have a strong understanding of quotient spaces until now. Thank you so much for the help, I worked through the details and it all makes sense! $\endgroup$ – The math god Oct 16 '18 at 0:42
  • $\begingroup$ @Themathgod If the answer solves completely your question, please mark it ss valid (click on the cutch symbol). $\endgroup$ – Dog_69 Oct 17 '18 at 23:12

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