Theorem 18.2 of Munkres’ Topology

Question

(b) (Inclusion) If $A$ is a subspace of $X$, the inclusion function $j:A\to X$ is continuous.

(c) (Composites) If $f: X\to Y$ and $g: Y\to Z$ are continuous, then the map $g\circ f:X\to Z$ is continuous.

(d) (Restricting the domain) If $f: X\to Y$ is continuous, and if $A$ is a subspace of $X$, then the restricted function $f_{A}:A \to Y$ is continuous.

(e) (Restricting or expanding the range) let $f: X\to Y$ be continuous. If $Z$ is a subspace of $Y$ containing the image set $f(X)$, then the function $g: X\to Z$ obtained by restricting the range of $f$ is continuous. If $Z$ is a space having $Y$ as a subspace, then the function $h:X\to Z$ obtained by expanding the range of $f$ is continuous.

(f) (Local formulation of continuity) the map $f:X\to Y$ is continuous if $X$ can be written as the union of open sets $U_\alpha $ such that $f_{U_\alpha}$ is continuous for each $\alpha$.

Motivation for this post is to prove all properties of functions that is used in Munkres proof of this theorem.

My attempt: (b) Claim: $j^{-1}(U)=A\cap U$. This one is easy. Proof: let $x\in j^{-1}(U)=\{ x\in A| j(x)\in U\}$. So $x\in A \wedge j(x)=x\in U$. $x\in A\cap U$. Conversely, let $x\in A \cap U$. So $x\in A \wedge x=j(x)\in U$(Since $x\in A$). Thus $x\in j^{-1}(U)$. Is inclusion map is same as identity map?

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(c) Claim: $(g\circ f)^{-1} (U)=f^{-1}(g^{-1}(U))$, where $U\subseteq Z$. Proof: let $x\in (g\circ f)^{-1} (U)$. Then $g(f(x))\in U$. The set $f^{-1}(g^{-1}(U))=\{ m\in X| f(m)\in g^{-1}(U)\}$. $g(f(x))\in U \Rightarrow f(x)\in g^{-1}(U)$. Thus $x\in f^{-1}(g^{-1}(U))$.

Conversely, suppose $x\in f^{-1}(g^{-1}(U))$. So $f(x)\in g^{-1}(U)=\{ y\in Y| g(y)\in U\}$. So $g(f(x))\in U$. Which implies $x\in (g\circ f)^{-1}(U)$. Hence $(g\circ f)^{-1}(U)=f^{-1}(g^{-1}(U))$. Is this proof correct?

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(d) Approach(1): let $V\in \mathcal{T}_{Y}$. Since $f$ is continuous, $f^{-1}(V)\in \mathcal{T}_{X}$. Claim: $f_A^{-1}(V)=A\cap f^{-1}(V)$. Proof: let $x\in f_A^{-1}(V)=\{x\in A| f_A (x)\in V\}$. Then $x\in A$ and $f_A(x)=f(x)\in V$(Since $x\in A$ and by definition of map $f_A$). Thus $x\in A \cap f^{-1}(V)$.

Conversely, let $x\in A \cap f^{-1}(V)$. So $x\in A$ and $f(x)=f_A(x)\in V$(Since $x\in A$, $f_A(x)=f(x)$). Thus $x\in f_A^{-1}(V)$. Is this proof correct?

Approach(2): Claim: $f_A = f\circ j_A$. Proof: $x\in A$. $f\circ j_A (x)= f(j_A(x))= f(x)=f_A (x)$(since $x\in A$).

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(e) Claim: $f^{-1}(U) \ = \ g^{-1}(B)$. Proof: https://math.stackexchange.com/a/3120401/861687.

Claim: $h=j_{Y}\circ f$. Proof: let $x\in X$. $j_{Y}\circ f(x)=j_{Y}(f(x))=f(x)=h(x)$, by definition of map $h$.

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(f) How to show $f^{-1}(V)\cap U_\alpha = f_{U_\alpha}^{-1}(V)$ and $f^{-1}(V)=\bigcup_{\alpha} (f^{-1}(V) \cap U_{\alpha})$?

Answer 1

As to (b): the inclusion map sort of is the identity map as $j(x)=x$ so as a "set of ordered pairs" it's the same as the identity. But in topology (and analyis etc) we use functions with a pre-specified domain and codomain so then there is a formal difference between the identity $1_A: A \to A$ and $j_A: A \to X$ when $A \subseteq X$; in the latter case the codomain is $X$ so to talk about continuity we need to consider open sets of $X$ and their inverse images etc. As you do. You see that the subspace topology is "designed" to be the minimal topology that makes $j_A$ continuous (it's a so-called initial topology). More on initial topologies and their universal property I wrote here. It's quite abstract, beware, but this is in essence how I was taught about subspaces and products and the like.

As to (c): in section 2, exercise 4(a) he already lets you show $(g\circ f)^{-1}[U]= f^{-1}[g^{-1}[U]]$ for all $U\subseteq Z$ (in other letters). Munkres assumes that all students do all exercises in linear order so this is a "known fact" for the model audience. So no need for a new proof here, easy though it is.

As to (d) just note that $f_A = f \circ j_A$, by definition, so (b)+(c) together imply its continuity. This is why he chose the order as he did.

As to (e): if $f[X]\subseteq Z \subseteq Y$ then the codomain restricted map $g$ (really $f: X \to Z$) is continuous as $j_Z \circ g = f$ and $f$ is continuous by the universal property of initial maps. While if $Y \subseteq Z$ the expanded $h$ (really $f: X \to Z$ again) is the composition $j_Y \circ f$ ( true by definition!) and thus continuous by (c).

That $f^{-1}[V] \cap U_\alpha = f_{U_\alpha}[V]$ is just like $(b)$ but for a non-inclusion map. Finally, using the the $U_\alpha$ form a cover : $$f^{-1}[V]= f^{-1}[V] \cap X = f^{-1}[V] \cap \left(\bigcup_\alpha U_\alpha \right) = \bigcup_\alpha (f^{-1}[V] \cap U_\alpha) \text{ (distributivity)}$$