# Determining if certian properties of a topological space pass to its image under a quotient map.

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.

Example. $X = [0, 1]$, $Y = S^1$, $p: X \to Y$ given by identifying the endpoints: $$p(x) = e^{2 \pi i x}.$$ Now, $\pi_1(X) = 1$, but $\pi_1(Y) \cong \Bbb{Z}$, generated by the identity map.
• Thanks for the example, but what is wrong with $\omega \simeq \text{const}_{pt}$ $\implies p \circ \omega \simeq p \circ \text{const}_{pt} = \text{const}_{p(pt)}$. Where exactly is the mistake? I thought it was a fact that $\omega \simeq \text{const}_{pt}$ $\implies p \circ \omega \simeq p \circ \text{const}_{pt}$, and the last equality just follows by the definition of a function (because $\text{const}_{pt}$ sends all points to a single point, and $p$ sends that point to exactly one point in Y), right? – Artus Jun 2 '14 at 1:13
• That argument is sound. That's what I was trying to say in the second paragraph ("...you have correctly observed that a homotopy between paths passes to the quotient...") The problem is that some loops in $Y$ lift to (are the image of) certain paths in $X$ that do not close. Hence, you can have a homotopy group generator in $Y$ that is not in the image of $p_* : \pi_1(X) \to \pi_1(Y)$. – Sammy Black Jun 2 '14 at 1:42