Let $X$ be a space equipped with topology $\mathcal{T}$. Determine the final topology on a set $Y$ induced by the constant maps $g:X \to Y$. 
Let $X$ be a space equipped with topology $\mathcal{T}$. Determine the final topology on a set $Y$ induced by the constant maps $g:X \to Y$.

If $g(x)=y_0$ for every $x \in X$ and the final topology is defined as $\mathcal{T}’ = \{U \subset Y : f^{-1}(U) \in \mathcal{T} \}$, then for any $f^{-1}(U) \in \mathcal{T}$ I have $f^{-1}(U) = \{x \in X \mid y_0 \in U \}$ and this set has to be in the topology on $X$. It seems that this is should be the discrete topology, but how to prove this from here?
 A: You are correct that the discrete topology is the answer. In fact, using the definition in the link, we can prove this without any heavy work or working with preimages.

*

*First note that if $g$ is continuous w.r.t. $(Y, \mathcal{T}_{\text{discrete}})$, then the answer is indeed "the discrete topology" since the discrete topology is finer than any topology on $Y$.

*$g$ is indeed continuous w.r.t. $(Y, \mathcal{T}_{\text{discrete}})$ because constant maps are always continuous!


Note. Here I have taken the definition to be:

The finest topology on $Y$ which makes $g$ continuous.

The description you've given is not what I've used (that description explicitly describes the topology). However, we can do it using that as well.
As you've written, we have $$\mathcal{T}’ = \{U \subset Y : f^{-1}(U) \in \mathcal{T}\}.$$
We now wish to show that every set $U \subset Y$ is open. Equivalently, we wish to show that $f^{-1}(U)$ is open (in $X$) for every subset $U \subset Y$.
But this is simple, because $f^{-1}(U)$ is either $X$ or $\varnothing$, depending on whether $U$ contains the image of $g$ or not. By definition of topology, we must have $\varnothing, X \in \mathcal{T}$.
