Techniques to Prove a Map Open or Closed

Question

I've been finding it hard to verify that given maps are open or closed in practice. For example, in his discussion of the Mobius bundle, Lee's Introduction to Smooth Manifolds says without justification that if $E$ is the quotient of $\mathbb{R}^2$ by the equivalence relation $(x,y) \sim (x',y')$ if and only if $(x',y') = (x + n, (-1)^n y)$ for some $n \in \mathbb{Z}$, and $q : \mathbb{R}^2 \to E$ is the quotient map, then the restriction of $q$ to $[0,1] \times \mathbb{R}$ is closed and therefore also a quotient map.

Since he doesn't go into much detail on this, I assume that verifying that the restriction is closed isn't hard, but I can't seem to find any way to do it without delving deep into the weeds and proving it straight from the definition. In general I have trouble proving that maps are open or closed; are there any tricks or overall methods I should be aware of?

Answer 1

The equivalence relation says $(x,y)\sim(x',y')$ iff $x'-x\in\Bbb{Z}$ and $y'=\pm y$ which implies $E=[0,1)\times(\Bbb{R}^+\cup\{0\})$.

Denote the restricted domain by $D=[0,1]\times\Bbb{R}$, then consider the restricted map $q^*:D\to E$ defined explicitly by,

$$ q^*(x,y)= \begin{cases} (x,|y|) &\text{ if }x\in[0,1)\\ (0,|y|) &\text{ if }x=1\\ \end{cases} $$

The coordinate function for $y$ is obviously closed.

Take a closed set $U\subset D$ then the $x$-coordinate of elements in $U$ must take values in $[a,b]\subset[0,1]$, while the $x$-coordinate for the image must take values in $[a,b]$ OR $[a,1)\cup\{0\}$, both of which situations are closed since $[a,1)\cup\{0\}=([a,1]\cap [0,1))\cup(\{0\}\cap [0,1))$ (subspace). Because this is true for every closed set in $D$, so $q^*$ is closed.

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Add one method (according to the note from hunter):

We can also prove $q:\Bbb{R}^2\to E$ is closed. Take a closed set $V\subset\Bbb{R}^2$ and denote the range for the $x$-coordinate of elements in $V$ by $V_x=[m,n]$. Then, $$q(V)_x= \begin{cases} [m_1,n_1]\subset[0,1) &\text{ if } \Bbb{Z}\cap[m,n]=\varnothing\\ [|m|-\lfloor|m|\rfloor,1)\cup[0,|n|-\lfloor|n|\rfloor] &\text{ if } \Bbb{Z}\cap[m,n]\neq\varnothing \end{cases} $$ So, $q$ is closed, and so does $q^*$ by restricting its domain, using the properties of a subspace.

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General tips:

This is hard to generalize a tip for every situations, but the ideas are similar. I'm sure that professors have deeper understandings and more methods than I do.

1. To prove a map is open(closed), we can test the basis or subbasis elements (or closed sets/ complement of open sets). Sometimes it's easy, if we're given a homeomorphism, then it is immediately open and closed.

2. However, continuity doesn't imply anything about closedness or openness, e.g. $$ f(x)= \begin{cases} x, & x\ge0\\ 0, & x\le0 \end{cases} $$ $f$ is clearly continuous, but not open since $(-2,2)\mapsto[0,2)$ given the euclidean topology. Another one is $g(x)=|x|$, it is closed but not open since $(-1,2)\mapsto[0,2)$. Besides that, there exists continuous maps that are not open nor closed.

But the two explicit maps mentioned above are open when we're given the lower-limit topology. So,

3. Be aware of the topology. Normally we use the euclidean topology when we refer to the topological space $\Bbb{R}^n$ without any further condition about its topology.

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I Hope this is useful.

Answer 2

While I'm aware that you asked for general methods, here is an elementary approach (partly because I haven't come across or figured out a straightforward abstract argument yet).

Let's denote the restriction of $q$ to $S=[0,1]\times\mathbb R$ by $q_S$, so that the symbol $q$ is reserved for the map with domain $\mathbb R^2$. Let $A\subset [0,1]\times\mathbb R$ be closed. We want to show that $q_S(A)=q(A)$ is closed. By definition of the quotient topology, $q(A)$ is closed iff $$ q^{-1}(q(A))=\{(x+n,(-1)^n y):(x,y)\in A\} $$ is closed. Since $\mathbb R^2$ is first countable, checking that $q^{-1}(q(A))$ is closed is equivalent to checking that for each convergent sequence $(w_k)_k$ in $q^{-1}(q(A))$ its limit $w$ lies in $q^{-1}(q(A))$. By the description above we can write $w_k=(x_k+n_k,(-1)^{n_k}y_k$) with $(x_k,y_k)\in A$. We have $w=(x,y)\in [m,m+1]\times [n,n+1]$ for some $m,n\in\mathbb Z$.

We consider two cases.

Case 1: If $x\in (m,m+1)$ then we know that eventually $w_k=(x_k+m,(-1)^my_k)$ for each $k$, and hence $(x_k,y_k)$ converges to $(x-m,(-1)^m y)\in A$ since $A\subset S$ is closed, and therefore $w=(x,y)\in q^{-1}(q(A))$.

Case 2: If $x=m$ or $x=m+1$, then we know that there exists a subsequence $x_{k_i}$ of $x_k$ such that either $x_{k_i}=0$ for all $i$ or $x_{k_i}=1$ for all $i$. In either case we have $(x_{k_i}+n_{k_i},(-1)^{n_{k_i}}y_{k_i})=(x_{k_i}+m',(-1)^{m'}y_{k_i})$ for all $i$ (with $m'=m$ or $m'=m+1$). By a similar argument as above it follows that $w\in q^{-1}(q(A))$.