# Images Form a Topology?

I have a function $f:X\rightarrow Y$ between two topological spaces $(X,\mathcal{T}_1)$ and $(Y,\mathcal{T}_2)$ that is $\mathcal{T}_1$-$\mathcal{T}_2$ continuous, and I'm asked to determine if the set of images {$f(V):V\in \mathcal{T}_1$} is a topology on $Y$. I've been wracking my brain on this one for days, and can't find a counterexample, but can't find a formal proof either. Help would be appreciated!

The result is not true in general.

Let $X=(0,1)$ and $Y=\mathbb{R}$. Let $f\colon X\to Y$ be the inclusion map. For $\{F(V):V\in\mathcal{T}_{1}\}$ to be a topology on $\mathbb{R}$, we need $\mathbb{R}\in \{f(V):V\in\mathcal{T}_{1}\}$, which we don't have.

• However, if $f$ is onto, the statement is true. – lattice Jul 10 '17 at 23:28
• @lattice. For the inclusion map or in general? – William Elliot Jul 11 '17 at 2:34
• I guess I'm stuck on the terminology of "a topology on $Y$". So that means that $Y$ HAS to be the set? I guess I am confused, cause I'm thinking that ${{0},{1},{0,1},\emptyset$ is a topology on \mathbb{R}, but I guess it's more of a subspace topology, is that fair? – Craig Jul 11 '17 at 2:49
• @Craig. Y is a set. A topology for Y is a subset of the power set of Y. The crammed together and lacking a bracket set you defined is not a topology for R. Why? – William Elliot Jul 11 '17 at 3:03
• @lattice. See my answer where I give an counter example and a proposition. – William Elliot Jul 13 '17 at 20:26

If f:X -> Y is a continuous surjection, then T = { f(U) | U open }
may not be a topology for Y. Example:

X = [0,2], Y = S^1, f(x) = (sin pi.x, cos pi.x)
[0,1) and (1,2] are open, so f([0,1}) cap f((1,2]) = {(0,1)}
should be open but there is no open U for which f(U) is a
singleton.

However if f is a bijection, then T is a topology for Y.
The continuity of f is unimportant. Only that X is a
topological space and f is a bijection.

Exercise. If f is an injection, show f(A cap B) = f(A) cap f(B).