Injectivity class of connected spaces in Top

Question

Consider this one-line statement and its proof. All connected spaces do not form an injectivity class in Top. Proof: let $m:A\to A'$ be a continuous map such that every connected space is $\{m\}-$injective.Using sufficiently large connected space with all subspaces of cardinality $|A'|$ discrete it is easy to show that for each clopen $U\subseteq A$ there is a clopen $U'\subseteq A'$ with $U=m^{-1}(U')$. It follows that the two point discrete space is also $\{m\}-$injective.

Questions: I have several questions about this short proof which I cannot solve myself.

1. I do not follow how the two point discrete space was created at the end of this proof.

2. How large must be the "sufficiently large connected space" from the middle of the proof?

3. How did we use the cardinality $|A'|$ in "all subspaces of cardinality $|A'|$ discrete"?

4. Finally why did we consider this preimage $U=m^{-1}(U')$?

It appears that I do not understand everything in this proof except for what we are to prove.

Answer 1

1. I'm not sure what you mean. The two-point discrete space is just the space with two points in which every subset is open.
2. It must be large enough to be connected while every subspace of cardinality $|A'|$ is discrete. So, just rewrite this line to "Using a connected space with every subspace of cardinality $|A'|$ discrete..."
3. We use it in proving the claim about clopens. Given a clopen $U\subset A$ and a space $M$ as in the previous point, we can define a continuous map $t:A\to M$ by sending $U$ to any point $x\in M$ and $A\setminus U$ to any $y\neq x$, since $\{x,y\}\subset M$ is discrete by assumption (we need to have ruled out that $A'$ is empty or a point separately.) Now factor $t$ through $m$ via $t':A'\to M'\subset M$, where the image $M'$ of $t'$ is, by assumption, discrete and contains $x$. Thus $(t')^{-1}(\{x\})$ is clopen in $A'$, and its inverse image under $m$ is $U$, as desired.
4. Hopefully clarified by the previous point.

Answer 2

The proof is perhaps easier to understand when presented top-down. Actually, Kevins answer already contains all the technical details.

For a set $M$ of (continuous) maps, write $inj(M)$ for the class of spaces that are $m$-injective for every $m \in M$. We want to show:

(a) if all connected spaces are in $inj(M)$ then the discrete space $2 = \{0,1\}$ is also in $inj(M)$.

Because of $inj(M) = \bigcap \{inj(m)\mid m\in M\}$ it is enough to consider the case of a single map $m\colon A \to A'$ and prove the following statement:

(b) if all connected spaces are $m$-injective then the discrete space $2 = \{0,1\}$ is also $m$-injective.

For a space $X$, write ${\cal C}(X,2)$ for the set of continuous maps from $X$ to $2$ and $clopen(X)$ for the set of clopen subsets of $X$. Then the following three conditions are equivalent:

(i) $2$ is $m$-injective.

(ii) $ {\cal C}(m,2) \colon {\cal C}(A',2) \to {\cal C}(A,2) $ is surjective.

(iii) $ m^{-1} \colon clopen(A') \to clopen(A) $ is surjective.

Here ${\cal C}(m,2)$ is precomposition with $m$, i.e. it maps $h \in {\cal C}(A',2)$ to $h\circ m \in {\cal C}(A,2)$ and $m^{-1}$ maps $U \in clopen(A')$ to $m^{-1}(U) \in clopen(A)$.

The equivalence (i) $\iff$ (ii) is just a rephrasing of the condition for $m$-injectivity: for each $f\colon A\to 2$ there exist a $g\colon A'\to 2$ with $f=g\circ m$.

In order to show (ii) $\iff$ (iii) observe that the correspondence given by $${\cal C}(X,2) \ni f \mapsto f^{-1}(1) \in clopen(X)$$ $$ clopen(X) \ni U \mapsto char_U \in {\cal C}(X,2) $$ is in fact natural in $X$. In particular we have:

$$ \matrix{ & {\cal C}(A',2) & \cong & clopen(A') & \cr {\cal C}(m,2) & \downarrow & & \downarrow & m^{-1} \cr & {\cal C}(A,2) & \cong & clopen(A) & \cr } $$

Therefore it is enough to prove the following statement:

(c) if all connected spaces are $m$-injective then $ m^{-1} \colon clopen(A') \to clopen(A) $ is surjective.

The rest is now as in Kevin's answer:

(0) for $A=\emptyset$ (c) holds because $\emptyset = m^{-1}(\emptyset)$. So we may assume $A\neq \emptyset$.

(1) take a (infinite) regular cardinal $\lambda > |A'|$ and a set $X$ with $|X|=\lambda$ equipped with a topology where the closed subsets (beside $X$) are those with cardinality $<\lambda$. If $W \subseteq X$ with $|W|<\lambda$ and $w\in W$, then $\{w\} = W \cap (X \setminus (W \setminus \{w\})$ is open in the subspace topology of $W$, so $W$ is discrete. Also $X = W \cup (X \setminus W)$ ensures that $W$ and $X\setminus W$ cannot both have cardinality $<\lambda$ because $\lambda$ is regular. So $X$ has no nontrivial clopen subsets and hence is connected.

(2) Given $U\in clopen(A)$, take $x,y\in X$ with $x\neq y$ and define $f\colon A \to X$ by $$ f(a) = \cases{ x & , if $a\in U$ \cr y & , if $a \notin U$ } $$

(3) Let $g\colon A' \to X$ with $f=g\circ m$ and set $U'=g^{-1}(x)$. Then $ U=f^{-1}(x) = m^{-1}(g^{-1}(x)) = m^{-1}(U')$. Because $|g(A')| \leq |A'| < \lambda $ the image $g(A')$ is discrete and $U' = g^{-1}(x) \in clopen(A')$.