# A universal property for the subspace topology

Let $X$ be topological space and $Y$ be a subset of $X$ with $i\colon Y\to X$ the inclusion map. Show that the induced topology of $Y$ is characterized by the following property: A function $f\colon Z \to Y$ of a topological space $Z$ into $Y$ is continuous if and only if $i\circ f$ is continuous.

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I wasn't sure what "application" meant and I know what the property should be, so I edited the question. Please consider writing about what you've tried, and phrasing the question in such a way that it doesn't sound like you're ordering folks around. [Textbooks are allowed to do this, but I doubt you'd walk into a colleague's office and say exactly what you wrote above; that would be somewhat rude.] Welcome, by the way! –  Dylan Moreland Jun 22 '12 at 1:32
@DylanMoreland "application" is French for mapping; a translator's false friend that often trips up French-speaking people on English-language math websites. –  user31373 Jun 22 '12 at 1:46
It is also helpful to see this definition of induced topology on a subset as a special case of the notion of initial topology with respect to a set of partial functions $f_\lambda: X \to X_\lambda, \lambda \in \Lambda$, where $X$ is a set, and $X_\lambda, \lambda \in \Lambda$ is a family of topological spaces. This is the topology which has a subbase the sets $f^{-1}_\lambda(U)$ for all open $U$ in $X_\lambda$ and $\lambda \in \Lambda$. It has the universal property that a function $f: Y \to X$, where $Y$ is a topological space, is continuous if and only if $f_\lambda \circ f$ is continuous for all $\lambda$. This is how for example one defines the topology on a manifold when the $f_\lambda$ are a family of charts.