# Find the equivalence class of $X\coprod Y$

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!

• What map is $i_Y$? – Dog_69 Oct 15 '18 at 22:18
• My apologies, it's just the inclusion mapping associated to $Y$. I edited the post. – The math god Oct 15 '18 at 22:19
• 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 $X$? – Dog_69 Oct 15 '18 at 22:22
• Actually, I am pretty sure you're right. They should be inclusions into the disjoint union. My apologies. – The math god Oct 15 '18 at 22:26
• Don't worry, it's okay. Now, you should be able to solve cases 1 and 3. – Dog_69 Oct 15 '18 at 22:30

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 $$f^{-1}(\, f(d)\,)$$.

• 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! – The math god Oct 15 '18 at 23:23
• 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 $D$.) – Berci Oct 15 '18 at 23:42
• 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! – The math god Oct 16 '18 at 0:42
• @Themathgod If the answer solves completely your question, please mark it ss valid (click on the cutch symbol). – Dog_69 Oct 17 '18 at 23:12