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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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    @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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    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