Let $X$ be a topological space, $A \subset X$ and $f: Z\longrightarrow A$ a function. Is it true that $f$ is continuous if and only if the function $i \circ f$ is also continuous (where $i: A \hookrightarrow X$ is the insertion of $A$ into $X$, i.e. $i(x)=x \ \ \ \forall x \in A$)?
This is a homework in my topology class, and I couldn't figure it out. Can anyone help?
It is easy to show that the inclusion and composites of continuous maps are continuous. Once you have established those (or found reference in your course material), it is easy to show that the continuity of $f$ implies that of $f\circ i$. For the converse, consider the inclusion $i\colon \mathbb{Q}\to\mathbb{R}$ and the characteristic function $\chi_\mathbb{Q}\colon\mathbb{R}\to \{0, 1\}$ where $\chi_\mathbb{Q}(x)=1$ if $x\in\mathbb{Q}$ and $\chi_\mathbb{Q}(x)=0$ otherwise.
$i \circ f$ is not defined when $i : A \to X$ and $f: Z \to X$, as the codomain of $f$ is not a always subset of the domain of $i$.
But when it is defined (i.e. when $f[Z] \subseteq A$) it will be continuous as a composition of continuous functions.
What is true is that
a function $f:Z \to A$ is continuous iff $ i \circ f: Z \to X$ is continuous.
because $A$ has the initial topology w.r.t. $i$. For detailed definitions and proofs see my answer here.
Quick proof for this special case: if $f$ is continuous so is $i \circ f$ (composition). OTOH, if $i \circ f$ is continuous, let $O$ be open in $A$, i.e. $O= O' \cap A$ for some open $O' \subseteq X$.
Observe that $O' \cap A= i^{-1}[O']$ and the given continuity implies that $(i \circ f)^{-1}[O']$ is open in $Z$ and that set equals $f^{-1}[i^{-1}[O']] = f^{-1}[O]$ and so $f$ is continuous.
