A property $P$ of topological spaces is said to "pass to quotients" if whenever $p : X \rightarrow Y$ is a quotient map and $X$ has property $P$ then $Y$ has property $P$. For the following properties determine if it "passes to quotients" or give a counter example.
a. Compactness
Compactness passes to quotients, because $p$ is continuous and the image of a compact space (which is $Y$ in this case as $p$ is surjective) is compact.
b. Simply connectedness
The continuous image of a connected space is connected, so $Y$ is also connected. To show that it is simply connected, we need to prove that $\pi_1(Y)=1$. However, $p_{*}: \pi_1(X) \rightarrow \pi_1(Y)$ is defined by $[\omega] \mapsto [p \circ \omega]$. Since $\omega \simeq \text{const}_{pt}$ $\implies p \circ \omega \simeq p \circ \text{const}_{pt} = \text{const}_{p(pt)}$, we have $\pi_1(Y)=1$. So $Y$ is also simply connected.
c. Path connectedness
For any $a,b \in Y$, we have $a' , b' \in X$ such that $p(a') = a$ and $p(b')=b$. Since $X$ is path connected, there exists $\gamma : [0,1] \rightarrow X$ with $\gamma(0) = a'$ and $\gamma(1) = b'$. Hence, we have $p \circ \gamma : [0,1] \rightarrow Y$ with $p \circ \gamma(0) = p(a') = a$ and $p \circ \gamma(1) = p(b') = b$. Since $p$ and $\gamma$ are continuous, so is their composition.
d. Discreteness
I'm not really sure if I understood this one. Does it mean if the set $X$ has the discrete topology, then so must $Y$? If so, then let $a \in Y$. Since $X$ has the discrete topology, $p^{-1}(a)$ is open in $X$. By the definition of a quotient map, $a$ must be open in $Y$. Hence, $Y$ has the discrete topology.
Are my answers correct?
