- Final topology.
Let $I := \{1, \ldots, n\}$, $Y$ a set, $(X_i)_{i \in I}$ a collection of topological spaces, and $(f_i:X_i \to Y)_{i\in I}$ a collection of maps. Let $X := \prod_{i \in I} X_i$. We define a map $f:X \to Y^n$ by $f(x) = (f_1(x_1), \ldots, f_n (x_n))$ with $x = (x_1, \ldots, x_n)$. Let $\tau$ be the finest topology on $Y$ such that $f_i$ is continuous for all $i \in I$. Then $\tau$ is also the finest topology on $Y$ such that $f$ is continuous. Here $X$ and $Y^n$ are endowed with the product topology.
- Initial topology.
Let $I := \{1, \ldots, n\}$, $X$ a set, $(Y_i)_{i \in I}$ a collection of topological spaces, and $(f_i:X \to Y_i)_{i\in I}$ a collection of maps. Let $Y := \prod_{i \in I} Y_i$. We define a map $f:X \to Y^n$ by $f(x) = (f_1(x), \ldots, f_n (x))$. Let $\tau$ be the coarsest topology on $X$ such that $f_i$ is continuous for all $i \in I$. Then $\tau$ is also the coarsest topology on $X$ such that $f$ is continuous. Here $Y^n$ is endowed with the product topology.
I guess above statements are correct because, in product topology,
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The projection maps are continuous.
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A multi-valued map is continuous if and only if its components are continuous.
I would like to ask if below relation is correct.
Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces and $f:X \to Y$. Then below statements are equivalent.
- Given $\tau_X$, $\tau_Y$ is the finest topology on $Y$ such that $f$ is continuous.
- Given $\tau_Y$, $\tau_X$ is the coarsest topology on $X$ such that $f$ is continuous.
Best Answer
No. Let $Y$ be a one-point set. Then $Y$ admits exactly one topology, and any function $f\colon X\to Y$ is continuous, no matter what $\tau_X$ is.