Injectivity class of connected spaces in Top 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.


*

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

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

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

*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.
 A: *

*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.

*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..."

*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.

*Hopefully clarified by the previous point.

A: 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 $U' = g^{-1}(x) \in clopen(A')$.
