0.1  (§18) Continuous functions

• $\mathrm{★}$ Def. $f:X\to Y$ is continuous if $f^{-1}(V)$ is open in $X$ for every open $V\subset Y$. A continuous function is called a map.

  Note. Refer back to §2: preimages respect $\cup$, $\cap$, $-$. Images do not. That is the entire reason the definition takes this form.

• Lem. Composites of maps are maps.

  Proof. ${(g\circ f)}^{-1}(W)=f^{-1}(g^{-1}(W))$.

• $\mathrm{★}$ Lem. It is enough to test a basis: $f$ continuous $\iff f^{-1}(B)$ open for all $B\in ℬ$. And enough to test a subbasis: $\iff f^{-1}(S)$ open for all $S\in 𝒮$.

  Proof. Because $f^{-1}$ commutes with unions (basis) and with finite intersections (subbasis).

• Ex. Metric spaces: continuity $\iff$ for all $x$ and $\epsilon >0$ there is $\delta >0$ with . The $\epsilon$–$\delta$ definition, recovered.

• $\mathrm{★}$ Ex. $\mathrm{id}:ℝ\to {ℝ}_{\mathrm{\ell }}$ is not continuous ($[a,b)$ is not open in $ℝ$); $\mathrm{id}:{ℝ}_{\mathrm{\ell }}\to ℝ$ is continuous.

• Def. $f$ is continuous at $x$ if for each neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ with $f(U)\subset V$.

• Thm. $f$ continuous $\iff$ $f$ continuous at every $x\in X$.

  Proof. $\Leftarrow$ is the gluing trick again: $f^{-1}(V)={\bigcup }_{x\in f^{-1}(V)}{U}_x$.

• $\mathrm{★}$ Thm. TFAE:

  1. $f$ is continuous;

  2. $f(\overline{A})\subset \overline{f(A)}$ for every $A\subset X$;

  3. $f^{-1}(B)$ is closed for every closed $B\subset Y$.

  Proof. $(1)\Rightarrow (2)$ via the neighborhood criterion for closure; $(2)\Rightarrow (3)$ apply to $A=f^{-1}(B)$; $(3)\Rightarrow (1)$ complement.

• $\mathrm{★}$ Def. $f:X\to Y$ is a homeomorphism if it is bijective and both $f$ and $f^{-1}$ are continuous. Then $X\cong Y$, topologically equivalent.

• Lem. $\cong$ is an equivalence relation (identity, inverse, composite).

• Lem. For $f$ bijective, TFAE: $f$ homeomorphism; $U$ open in $X\iff f(U)$ open in $Y$; $V$ open in $Y\iff f^{-1}(V)$ open in $X$.

  Note. So a homeomorphism is a bijection ${𝒯}_X\to {𝒯}_Y$ of the topologies, not just of the points.

• Def. A topological property is one expressible in terms of points and open sets; such properties are preserved by homeomorphism. E.g. finiteness, discreteness, Hausdorff, and later compactness and connectedness.

• Ex. $(0,1)\cong (a,b)$ via $f(x)=(1-x)a+xb$.

• Ex. $(-1,1)\cong ℝ$ via $f(x)=x/(1-x^2)$,   $f^{-1}(y)=2y/\left(1+\sqrt{1+4y^2}\right)$. Hence every open interval $\cong ℝ$.

• $\mathrm{★}$ Caution. Continuous bijection $\ne$ homeomorphism.

  • $\mathrm{id}:{ℝ}_{\mathrm{\ell }}\to ℝ$.

  • $f:[0,1)\to S^1$, $f(t)=(\cos 2\pi t,\sin 2\pi t)$. Here $U=[0,\frac{1}{2})$ is open in $[0,1)$ but $f(U)$ is not open in $S^1$: no neighborhood of $(1,0)$ lies inside it.

Constructing maps.

• Thm. The subspace topology on $A\subset X$ is the coarsest making the inclusion $i:A\to X$ continuous.

• Cor. Restrictions $f|A$ of maps are maps. Lem. Corestrictions $X\to B$ (with $f(X)\subset B$) are maps.

• Def. $f:X\to Y$ is an embedding if the corestriction $X\to f(X)$ is a homeomorphism onto the subspace $f(X)$.

• Lem. $f$ is an embedding $\iff$ $f=j\circ h$ with $h$ a homeomorphism onto a subspace and $j$ the inclusion. Embeddings are injective.

• Ex. $[0,1)\to {ℝ}^2$, $t\mapsto (\cos 2\pi t,\sin 2\pi t)$: injective and continuous, not an embedding.

• Lem. Maps to a one-point space are continuous; hence constant maps are continuous.

• $\mathrm{★}$ Thm. (Pasting lemma.)

  • $X={\bigcup }_{\alpha \in J}{U}_{\alpha }$ with all $U_{\alpha }$ open: $f$ continuous $\iff$ every $f|U_{\alpha }$ continuous.

  • $X=A_1\cup \mathrm{\cdots }\cup A_n$ with all $A_i$ closed (finitely many): $f$ continuous $\iff$ every $f|A_i$ continuous.

  Proof. Open case: $f^{-1}(V)=\bigcup {(f|U_{\alpha })}^{-1}(V)$, a union of open sets. Closed case: same with closed sets — and now finiteness is essential.

• $\mathrm{★}$ Thm. (Maps into a product.) $f:A\to X\times Y$ is continuous $\iff$ both components $f_1={\pi }_{1}f$ and $f_2={\pi }_{2}f$ are continuous.

  Proof. Subbasis criterion: $f^{-1}(U\times Y)=f_1^{-1}(U)$, $f^{-1}(X\times V)=f_2^{-1}(V)$.

• Cor. The product topology is the coarsest topology making both projections continuous.

0.2  (§19) The product topology (general)

• $\mathrm{★}$ Def. On ${\prod }_{\alpha \in J}{X}_{\alpha }$ the product topology is generated by the subbasis $𝒮=\{{\pi }_{\beta }^{-1}(U_{\beta })\mid \beta \in J,U_{\beta }\subset X_{\beta }\text{ open}\}.$

• $\mathrm{★}$ Lem. Concretely:

  • subbasis elements are ${\prod }_{\alpha }{U}_{\alpha }$ with $U_{\alpha }\ne X_{\alpha }$ for at most one $\alpha$;

  • basis elements are ${\prod }_{\alpha }{U}_{\alpha }$ with $U_{\alpha }\ne X_{\alpha }$ for finitely many $\alpha$.

  Note. Emphasise “all but finitely many coordinates unrestricted”. This is the point where students expect the naive definition and get a different one.

• Thm. $f:A\to \prod X_{\alpha }$ continuous $\iff$ every component $f_{\beta }={\pi }_{\beta }f$ is continuous.

• Ex. [Mk §19 Ex. 2] (Product vs. box.) The box topology on $\prod X_{\alpha }$ has basis all ${\prod }_{\alpha }{U}_{\alpha }$ with every $U_{\alpha }$ open — no finiteness condition. Let $f:ℝ\to {ℝ}^{\omega }$, $f(t)=(t,t,t,\mathrm{\dots })$. Each $f_n(t)=t$ is continuous, so $f$ is continuous for the product topology — but not for the box topology.

  Proof. Take the box $B=(-1,1)\times (-\frac{1}{2},\frac{1}{2})\times (-\frac{1}{3},\frac{1}{3})\times \mathrm{\cdots }$. If $f^{-1}(B)$ were open, it would contain some $(-\delta ,\delta )\ni 0$; applying ${\pi }_n$ gives $(-\delta ,\delta )\subset (-\frac{1}{n},\frac{1}{n})$ for all $n$, impossible. In fact $f^{-1}(B)=\{0\}$.

  Note. So the theorem above fails for the box topology. Box is finer than product (they agree for finite products), and the more open sets the target has, the harder it is for a map into it to be continuous. This is why the product topology is the right one.

• Cor. The product topology is the coarsest making all ${\pi }_{\beta }$ continuous.

• Thm. $A_{\alpha }\subset X_{\alpha }$ subspaces: on $\prod A_{\alpha }$, product topology $=$ subspace topology from $\prod X_{\alpha }$.

• Thm. All $X_{\alpha }$ Hausdorff $\Rightarrow \prod X_{\alpha }$ Hausdorff.

• Thm. $\overline{{\prod }_{\alpha }{A}_{\alpha }}={\prod }_{\alpha }\overline{A_{\alpha }}$.