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?


Your answers a, c, and d are correct. However, b is not.

A space is simply connected if it's fundamental group is trivial. While you have correctly observed that a homotopy between paths passes to the quotient, it may be that certain open paths become closed loops in the quotient, hence generators in the fundamental group of the quotient space.

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.

  • $\begingroup$ 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? $\endgroup$ – Artus Jun 2 '14 at 1:13
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    $\begingroup$ 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 $p_* : \pi_1(X) \to \pi_1(Y)$. $\endgroup$ – Sammy Black Jun 2 '14 at 1:42

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