Do all continuous real-valued functions determine the topology? Let $X$ be a topological space. If I know all the continuous functions from $X$ to $\mathbb R$, will the topology on $X$ be determined?
I know the $\mathbb R$ here is somewhat artificial. So if this is wrong, will it be right if $X$ is a topological manifold?
 A: Spaces for which this is true are called completely regular. In fact, it is an equivalent characterization of completely regular spaces $X$ that their topology is entirely determined by the set $C(X)$ of real-valued continuous functions on them. In other words, there is a unique completely regular topology that makes all these and only these functions continuous. The standard definition however is this: given any closed subset $F$ and any point $x\not\in F$ there is a continuous real-valued function $f$ on $X$, which is constant $1$ on $F$ but $f(x)=0$. Any regular normal space is completely regular by the Urysohn's lemma, and topological manifolds are certainly completely regular.
A: Conifold's answer is essentially correct, but you should be careful what you mean by "determine the topology".  Given any collection $I$ of real-valued functions on a set $X$, there is a natural topology you can impose on $X$, namely the coarsest topology that makes each element of $I$ continuous.  Explicitly, this topology on $X$ is the collection of unions of finite intersections of sets of the form $f^{-1}(U)$ where $U\subseteq\mathbb{R}$ is open and $f\in I$.
Completely regular spaces are exactly the spaces $X$ such that their topology coincides with this natural topology induced by the set $I$ of all continuous real-valued functions on $X$.  In particular, if you know a space is completely regular, then you can canonically recover its topology from the set of all continuous real-valued functions on it (so in this sense they "determine the topology").  However, this does not mean that its topology is the only possible topology on the set with the same collection of continuous real-valued functions.
For instance, let $T$ be the usual topology on $[0,1]$, and let $T'$ be the topology on $[0,1]$ consisting of sets of the form $U\setminus A$ where $U\in T$ and $A\subseteq\{1,1/2,1/3,\dots\}$ (intuitively, think of $T'$ as the usual topology modified so that the sequence $(1/n)$ no longer converges to $0$).  Then a function $[0,1]\to\mathbb{R}$ is continuous with respect to $T$ iff it is continuous with respect to $T'$.  Indeed, suppose $f:[0,1]\to\mathbb{R}$ is continuous with respect to $T'$.  Then for any open $V\subseteq\mathbb{R}$ and any point $x\in f^{-1}(V)$ there is a neighborhood $W$ of $x$ such that $\overline{W}\subseteq f^{-1}(V)$, because $V$ contains a closed neighborhood of $f(x)$.  But it is not hard to show that an open set $U\in T'$ contains such a $W$ around each of its points iff actually $U\in T$.  It follows that $f$ is continuous with respect to $T$, not just with respect to $T'$.  (The other implication is trivial, since $T\subset T'$.)
Thus in this example, even though the topology $T$ is completely regular, there is still another topology $T'$ with the same continuous real-valued functions.  All that complete regularity guarantees you is that for any such topology $T'$, $T\subseteq T'$.  (It also guarantees you that $T'$ cannot also be completely regular unless $T=T'$, so $T'$ must be somewhat pathological.)
