Is $f:A \to f(A)$ continuous if $f :A \to \mathbb{R}$ is continuous? Let A be a closed interval. Suppose $f:A \to\mathbb{R}$ is continuous. Does that imply that $f:A \to f(A)$ is continuous? Note that $f(A) \subseteq \mathbb{R}$ is a subspace topology.
I am asked to prove $f:A \to f(A)$ is continuous. I do not understand why that requires proof. Does not it follow from $f:A \to\mathbb{R}$ being continuous? What does subspace topology have to do with this
 A: The topology on $f(A)$ is certainly important. The identity map $f$ from $\mathbb R$ to itself is continuous but as a map from $\mathbb R$ to $f(A)$ with the indiscrete topology it is not continuous. 
(The proof when $f(A)$ has subspace topology is in the comment above). 
A: Saying that $f: A \to \mathbb{R}$ is continuous (in the topological sense) depends on the topology on the domain space ($A$ in the subspace topology inherited from  $\Bbb R$) and the codomain space $\Bbb R$. So claiming that $f$ considered as a map from $A$ to $f[A] \subseteq \Bbb R$ is continuous is a claim that needs some thought, as we changed the codomain of the function and that has its own topology but one it inherits from $\Bbb R$, and the latter helps: if $O \subseteq f[A]$ is open this means that we can write it as $O' \cap f[A]$ for some open subset $O'$ of $\Bbb R$ (this is the definition of the subspace topology) and then $f^{-1}[O] = f^{-1}[O']$ and the latter is open in $A$ by the fact that our original $f: A \to \mathbb{R}$ is continuous. So $f:A \to f[A]$ is also continuous.
<optional>
A more abstract topological view: $f[A]$ has the initial topology wrt the inclusion map $i: f[A] \to \Bbb R$. Letting $f': A \to f[A]$ be $f$ in the codomain restriction of $f:A \to \Bbb R$ we have that $f = f' \circ i$ and then the universal map property of the initial topology implies that $f'$ is continuous iff $f$ is. </optional>
Another way to see it in this case is that the $\varepsilon-\delta$ equivalent of defining the continuity does not mention the codomain space at all: 
$$\forall x \in A: \forall \varepsilon>0: \exists \delta >0: \forall y \in A: |x-y| < \delta \to |f(x)-f(y)| < \varepsilon$$
and because we use the same $|\cdot|$ distance on $f[A]$ as we do in $\Bbb R$, continuity of both views of $f$ amounts to exactly the same.
