Can a set be homeomorphic to a quotient map from itself? Making sense of a problem. I have a problem (Willard 9H.3) that seems to have some typos or mixed symbols. I'm studying independently and recognize I could be wrong -- can someone read this over for sensibility? 
The background, which was sensible:
Suppose $X_\alpha$ is a topological space and $f_\alpha$ is a map of $X_\alpha$ to a set Y, for each $\alpha \in A$. The strong topology coinduced by the maps $f_\alpha$ on Y consists of all sets U in Y such that $f_\alpha^{-1}(U)$ is open in $X_\alpha$ for all $\alpha \in A$. 
The family of maps $f_\alpha$ will be said to cover points of Y iff each $ y \in Y$ is in the image of some $f_\alpha$. 
Let X be the disjoint union of the spaces $X_\alpha$. If x and y are points of X, then $x \in X_\alpha$ and $y \in X_\beta$ for some choice of indices $\alpha$ and $\beta$. Define an equivalence relation on X as (x ~ y) iff $f_\alpha(x) = f_\beta(y)$, and denote the resulting quotient space by Z. 
The exercise, which seems like it has a few typos: 
Prove, if the maps $f_\alpha$ cover points of Y, then Y has the strong topology coinduced by them iff X is homeomorphic to the quotient space Z constructed above, under the maph h defined as follows: for $y \in X$, pick $\alpha \in A$ and $x \in X$ so that $f_\alpha (x) = y$ and then define $h(y) = [x]$
My proposed correction to the exercise, replacing X with Y in two places: 
Prove, if the maps $f_\alpha$ cover points of Y, then Y has the strong topology coinduced by them iff Y is homeomorphic to the quotient space Z constructed above, under the maph h defined as follows: for $y \in Y$, pick $\alpha \in A$ and $x \in X$ so that $f_\alpha (x) = y$ and then define $h(y) = [x]$
Before I move ahead and try to prove the wrong thing, could someone read over this?
 A: Your correction is quite correct: we want to characterise the space $Y$ in the strong topology by making a concrete model for it from $X$ and the proposed quotient map. It shows all strong topologies are just quotients of sums. The proposed homeomorphism is fine as well. 


*

*First show $h$ is well defined: what if we pick another $\alpha$ and $x$, do we get the same image?

*Show $h$ is continuous. Let $U$ be open in $Z$. See if $h^{-1}[U]$ is open in $Y$ in the strong topology, and there's a simple test for that. If $i_\alpha$ is the canonical embedding of $X_\alpha$ into $X$ then note that $q \circ i_\alpha = h \circ f_\alpha$ which will help.

*Show the inverse (define it and show it's well-defined) is continuous too, using that a map from a quotient is continuous iff the composition with the quotient map is continuous (from $X$ here). And a map from a sum is continuous iff.... (fill in details)

A: Here is a very slightly different take on the problem (which is easier to follow if you draw some commutative diagrams, but I'm not sure that's so easy here on StackExchange):
Let $i_\alpha \colon X_\alpha \to X$ be the canonical injections, and let $f \colon X \to Y$ be the unique (continuous) map such that $f_\alpha = f \circ i_\alpha$. For $x,y \in X$, let $x \sim y$ if and only if $f(x) = f(y)$ (this is the same equivalence relation as in Willard's formulation).
Putting $Z = X/{\sim}$ and denoting the quotient map $X \to Z$ by $q$, there is a unique (continuous) map $\tilde f \colon Z \to Y$ such that $f = \tilde f \circ q$. Namely, $\tilde f([x]) = f(x)$. Here $\tilde f$ plays the role of $h^{-1}$ in Willard's notation.
If the $f_\alpha$ cover points in $Y$, then $f$ (and hence $\tilde f$) is surjective, and $\tilde f$ is injective by construction. And it is already continuous.
Now assume that $\tilde f$ is a homeomorphism. Then $Y$ has the strong topology coinduced by $\tilde f$. (Obviously. But this is a useful way of putting it.) But $Z$ has the strong topology coinduced by $q$, and $X$ also has the strong topology coinduced by the $f_\alpha$. It follows that $Y$ has the strong topology coinduced by the compositions $\tilde f \circ q \circ i_\alpha$. (This fact is the dual of Willard's Exercise 8H.1, but even more obvious.) However,
$$ \tilde f \circ q \circ i_\alpha = f \circ i_\alpha = f_\alpha, $$
so $Y$ carries the strong topology coinduced by the $f_\alpha$ as desired.
Conversely, suppose that $Y$ has the strong topology coinduced by $f_\alpha$. It suffices to show that $\tilde f$ is an open map. To this end, let $U \subseteq Z$ be open. Then $\tilde f(U)$ is open iff $f_\alpha^{-1}(\tilde f(U)) = (q \circ i_\alpha)^{-1}(U)$ is open in $X_\alpha$ for all $\alpha$. But since $Z$ carries the strong topology coinduced by the $q \circ i_\alpha$, this is true.
