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If $h:A\to B$ is a homeomorphism, where the subset $A_1$ of $A$ is homeomorphic to a subset $B_1$ of $B$. How can I prove that the quotient spaces $A/A_1$ and $B/B_1$ are homeomorphic?

Thanks

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    Is $B_1 = h(A_1)$? Or are they just independently homeomorphic with the induced topologies?2012-11-18
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    @uncookedfalcon yes, $B_1=h(A_1)$2012-11-18
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    cheers, writing up answer2012-11-18
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    @uncookedfalcon thanks :)2012-11-18

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We are given maps $h: A \rightarrow B$, $g: B \rightarrow A$ such that $gh = id_A, hg = id_B$ (namely take $g = h^{-1}$), $h(A_1) = B_1$.

To give a map out of $A/A_1$ is to give a map out of $A$ which sends $A_1$ to a point: consider $A \xrightarrow{h} B \rightarrow B/B_1$, this induces $h': A/A_1 \rightarrow B/B_1$, similarly we get $g': B/B_1 \rightarrow A/A_1$, one observes that $h', g'$ are by construction inverse, as desired.

I think the claim is false if they're independently homeomorphic. Consider $A = B = [0,1]$, $A_1 = \{0,1\}, B_1 = \{1/3, 2/3\}$. Then $A/A_1$ is just $S^1$, but $B/B_1$ looks like:

enter image description here

Now if I pluck out $1/3 = 2/3$ from $B/B_1$ I get something with 3 connected components, but if you take out any point from $S^1$, you get just 1 connected component.

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    You use implicitly the additional hypothesis that $h$ restricts to a homeomorphism $A_1 \to B_1$. You correctly note that the conclusion is false without this hypothesis, but before you note that, it would be a good idea to actually state it...2012-11-18
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    A fair point-editing!2012-11-18
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    I made the above comment before reading the comments on the question. In the context of those comments this answer is less confusing, but in general I still think it's a good idea to write answers independent of the comments (unless you explicitly cite the comments as necessary, e.g. "as discussed in the comments we assume...").2012-11-18
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    Yeah that's reasonable man. Cheers!2012-11-18
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    What is exactly this induced map $h'$?2012-11-18
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    The quotient map $q: A \rightarrow A/A_1$ sends all of $A_1$ to a point. For any other map $f: A \rightarrow B$ sending $A_1$ to a point, we can factor uniquely it through $A/A_1$ as follows: take a point of $A/A_1$: if it's $a$, some point other than the distinguished point * gotten from mashing $A_1$ down to a point, send it to $f(a)$, if it's *, pick any $a_1 \in A_1$ and send * to $f(a_1)$. That's the induced map.2012-11-18
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    how can we prove that the induced map is continuous?2012-11-18
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    Take an open $U$ of $B$. We want to know that when we pull it back into $A/A_1$, it's still open. Well, it's open in $A/A_1$ iff when we further pull it back into $A$ it's open (by def of the quotient topology), and by assumption $A \rightarrow B$ was continuous2012-11-18
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    sorry, but this part of the explanation is a little bit confused: "if it's a, some point other than the distinguished point * gotten from mashing A1 down to a point, send it to f(a), if it's *, pick any a1∈A1 and send * to f(a1). That's the induced map."2012-11-18