Let $X$ be a set with an equivalence relation $\sim$. Let $f : X/\sim \ \to Y$ be a function. We define a function $\tilde{f}$, called the lift of $f$, as follows: $\tilde{f} : X \to Y$, $x \mapsto f([x])$.
(a) If $f$ is injective, then $\tilde{f}$ is injective. Prove or give a counterexample.
(b) If $f$ is surjective, then $\tilde{f}$ is surjective. Proof or give a counterexample.
My question here is, can I just assume that $\tilde{f} = f \circ \pi$, where $\pi: X \to X/ \sim$ is the quotient map? For (b), since $\pi$ is surjective and we assume $f$ is surjective, then $\tilde{f}$ is surjective. For (a), to show a counter example, I have to pick a specific $X,Y$, and function $f$ such that $\tilde{f}$ is not injective? So this:
Let $X = \mathbb{Z}$ and let $Y=\{0,1\}$. Define the relation $\sim$ by $a \sim b \iff a \equiv b \mod 2$ for $a,b \in \mathbb{Z}$. The equivalence classes are $[0]$ and $[1]$, so $X / \sim \ = \{[0], [1]\}$. Now define $f: X/ \sim \ \to Y$ by $[0] \mapsto 0$ and $[1] \mapsto 1$, and let $\tilde{f}$ be defined as above. Note that $f$ is clearly injective. Consider $1,3 \in \mathbb{Z}$. Of course, $1\ne 3$, but since $3 \equiv 1 \mod 2$, we see that $[1]=[3]$. So $$ \tilde{f}(1) = f([1])=f([3])= \tilde{f}(3). $$ Thus, $\tilde{f}$ is not injective even though $f$ is injective.
Also, what is a lift exactly?
Yes, you can assume that $$\overline{f} = f\circ \pi$$ since $(f\circ \pi)(x) = f(\pi(x)) = f([x]) = \overline{f}(x)$.
a) This is not true in general. We can prove it by contraposition. Assume $\overline{f}$ is not injective and $\sim$ is given by $x\sim x' :\Leftrightarrow \overline{f}(x) = \overline{f}(x')$. That is, we identify all elements of $X$ that have a common image in $Y$.
But even though $\overline{f}$ was not injective, $f$ is injective since $[x] \not= [x']$ now implies $f([x]) \not= f([x'])$.
b) Your thoughts on b) are correct.
In category theory a lift describes the following:
given a morphism $f\colon X\to Y$ and a morphism $g\colon Z \to Y$, a lift of $f$ to $Z$ is a morphism $h\colon X \to Z$ such that $$f = g\circ h.$$ That is, $f$ factors through $h$.
If you are not familiar with category theory, the morphisms in the category of topological spaces are the continous maps.