Let $f$ be a continuous mapping of a compact metric space $(X, d)$ onto a Hausdorff space $(Y, \tau_1)$. Then $(Y, \tau_1)$ is compact and metrizable.
In one proof the following metric is constructed:
$d_1(y_1, y_2) = inf\{d(a, b) : a \in f^{-1}\{y_1\},\ b \in f^{-1}\{y_2\}\}$, for all $y_1$ and $y_2$ in $Y$.
I'm thinking about how to prove that triangle inequality holds for $d_1$, s.t. $d_1(x, z) \leq d_1(x, y) + d_1(y, z)$
And how about the case when triangle inequality does not hold?
Update:
@daniel-schepler suggested to use metric $d_1(x, y) = inf\{d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)\}$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$.
Now we need to prove 3 property of metric: positivity, symmetry and triangle inequality. This is my sketch:
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Positivity. Chose two points $x,y \in Y$ s.t. $x \neq y$. Singleton sets $\{x\}$ and $\{y\}$ are disjoint closed sets by Hausdorfness of $Y$ and their preimages are also disjoint closed in $X$ by continuity of $f$. $X$ is compact metric space by hypothesis and therefore a normal space. In normal space every two disjoint closed sets have disjoint open neighborhoods. Let $f^{-1}(x) \subseteq U$ and $f^{-1}(y) \subseteq V$ s.t. $U,V$ are open and $U \cap V = \varnothing$. Therefore there exists an $\varepsilon > 0$, s.t. $x^* \in B_\varepsilon(x^*) \subseteq U$ and $y^* \in B_\varepsilon(y^*) \subseteq U$ and distance between any two closed sets is strictly positive and $d_1$ is positive and zero iff $x = y$.
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Symmetry. Clearly $d_1$ is symmetric.
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Triangle inequality. For triangle inequality we can show that in the "worst" case $d_1(x, z) = d_1(x,y) + d_1(y, z)$.
Best Answer
No, this does not form a metric in general. For a specific counterexample, consider the case $X = [0, 1]^2$, $Y = \{ (x, y) \in [0, 1]^2 \mid x \le y \}$, and $f : X \to Y$ is defined by $f(x, y) = (xy, y)$. Then for all points in $Y$ with $y \ne 0$, the inverse image of $(x, y)$ is a single point $\{ (x/y, y) \}$; whereas the inverse image of $(0, 0)$ is $[0, 1] \times \{ 0 \}$. Thus, for example, in the induced $d_1$ on $Y$, we have $d_1((0, 0.1), (0.1, 0.1)) = d((0, 0.1), (1, 0.1)) = 1$. On the other hand, $d_1((0, 0.1), (0, 0)) = 0.1$ and also $d_1((0, 0), (0.1, 0.1)) = 0.1$.
You could try to fix this by considering paths, e.g. define $d_1(x, y)$ to be the infimum of values of the form $d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$. Then it would be straightforward to show this does satisfy the triangle inequality. However, then the tricky part would be to show that this infimum is strictly positive for $x \ne y$. I'm not sure off the top of my head how (or even if) that could be proved, though it seems clear that compactness would have to come into play in any proof of such a fact.