What are some examples where the subspace topology is not the same as the topology defined directly on the subset? One example was given in Munkres. Take the order topology on $\mathbb{R}^2$ using the dictionary order. Then the subspace topology on the unit square is not the same as the "ordered square" on the same subset.
Are there any criterion for when the subspace topology always equals the topology defined on the subset, or when they are not equal? Also, what are some other examples where they are not equal?
 A: Yes. There is a theorem in Munkres that states that for a convex subset of an ordered set, the order topology on the subset coincides with the subspace topology. (Convex means that if $a$, $b$ are in the subset, so is $[a,b]$.) The ordered square is not a convex subset of $\mathbb R^2$, so the two topologies don't have to coincide. A simpler example is the set $A=[0,1)\cup\{2\}\subset \mathbb R$. The order topology gives that $A$ is homeomorphic to $[0,1]$, whereas in the subspace topology $\{2\}$ is an open set.
A: Let $\langle X,\tau\rangle$ be any topological space, and let $Y$ be any subset of $X$ with at least two points. Let $\tau_Y$ be the subspace topology on $Y$. Then we can always find a topology $\tau'$ on $Y$ such that $\tau'\ne\tau_Y$. A very simple way to do it is to notice that if $|Y|\ge 2$, the discrete and indiscrete topologies on $Y$ are distinct, so at least one of them must be different from $\tau_Y$.
If we already have some topology $\tau'$ on $Y$, the only time that we can be certain (without actually knowing what the topologies are) that $\tau'=\tau_Y$ is when $|Y|=1$.
A: I suppose you are thinking of something along the lines of the following:
Take a space $(X,\tau)$ with an equivalence relation $\sim$. If $(A,\tau_A)$ is a subspace of $X$, then we can restrict the relation $\sim$ to the relation $\sim_A$ on $A$. On the quotient set $X/\sim$, the set of all equivalence classes, the topology consists of all sets $U$ such that $q^{-1}(U)$ is open in $X$, where $q:X\to X/\sim$ is the quotient map.
Now, the most natural guess for the topology on $A/\sim_A$, the set of all $\sim_A$-equivalence classes in $A$ (which is in a natural way equivalent to the set $q(A)$, the set of all $\sim$-equivalence classes in $X$ with at least on point from $A$), would be the quotient topology again. But: this quotient topology on $A/\sim_A$ is not always the same as the topology inherited from $X/\sim$ (it is always finer, though).
To put it briefly

the quotient topology on the subspace is finer than the subspace topology of the quotient.

Or we can formulate it

The restriction of a quotient map need not be a quotient map .

Here is an example: Let $X=[0,1]\times[0,1]$ with the usual topology. Let $Y=X/B$, the space obtained by collapsing the lower edge $B:=[0,1]\times\{0\}$ to a point, and $q:X\to Y$ the quotient map. Let $A=[0,1]\times(0,1]\ \cup\ \{(1,0)\}$. Now the set $U=(1/2,1]\times I\cap A$ is open in $q(A)$, but no open set in $X/\sim$ can intersect $q(A)$ in $U$ because its preimage had to be an open set containing $I\times\{0\}$.
See also my answer in Finding a counterexample; quotient maps and subspaces for another counterexample, as well as Ronnie Brown's answer for a criterion for equality of the topologies.
