Topology – Every injective quotient map is a homeomorphism

general-topologylogicsolution-verification

I am going to show that every injective quotient map is a homeomorphism:

Let $(X,\tau_{X})$, $(Y,\tau_{Y})$ be topological spaces.

Definitions:

$q:X \rightarrow Y$ is a quotient map iff $q$ is surjective ($q[X] = Y$) and
$$
\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X
$$
where $[]$ is used to denote the image of a function. $f:X \rightarrow Y$ is a homeomorphism iff $f$ is bijective and
$$
\forall U\subseteq X: U\in \tau_{X} \iff f[U] \in \tau_Y
$$

Lemma:
$$\forall x \forall y: P(x,y) \implies Q(x)$$
is equivalent to
$$\forall x: Q(x) \land \forall x \exists y: P(x,y)$$

Proof of the lemma: proof

Proof:

It is sufficient to show that if $q$ is injective, $\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X$ is equivalent to $\forall U\subseteq X: U\in \tau_{X} \iff q[U] \in \tau_Y$.

Notes:
Injectivity of $q$ ensures $q^{-1}[q[U]] = U$ for all $U \subseteq X$.
For a surjection $q$, $\forall V \subseteq Y: \exists U \subseteq X: q[U] = V$ is a logical necessity.

$$
\begin{align}
&\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X& \\
&\iff
(\forall V\subseteq Y: V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X) \land (\forall V \subseteq Y: \exists U \subseteq X: q[U] = V)& \text{Tautology}\\
&\iff
\forall V\subseteq Y : \forall U \subseteq X: q[U] = V \implies V\in \tau_{Y} \iff q^{-1}[V] \in \tau_X & \text{Lemma}\\
&\iff
\forall U \subseteq X: \forall V\subseteq Y : q[U] = V \implies q[U]\in \tau_{Y} \iff q^{-1}[q[U]] \in \tau_X &p \rightarrow q \iff p\rightarrow p \land q\\
&\iff
\forall U \subseteq X: \forall V\subseteq Y : q[U] = V \implies q[U]\in \tau_{Y} \iff U \in \tau_X & \text{Injectivity}\\
&\iff
(\forall U\subseteq X : q[U]\in \tau_{Y} \iff U \in \tau_X) \land (\forall U \subseteq X:\exists V \subseteq Y: q[U] = V) &\text{Lemma}\\
&\iff
\forall U\subseteq X : q[U]\in \tau_{Y} \iff U \in \tau_X& \text{Tautology}\\
\end{align}
$$

Is this correct?

Best Answer

Your proof is correct, but it is difficult to understand and unnecessarily complicated.

Your lemma is a purely logical statement, and it is unusual to begin a concrete proof with such a statement. In fact, all arguments are based on logic, and we must take that foundation as granted. Otherwise every proof would have to begin with an explanation of all logical, set theoretical etc. issues.

The essence of your arguments it this:

  1. Any quotient map $q$ is a continuous surjection.

  2. Any injective quotient map $q$ is a continuous bijection. Thus it suffices to show that if $U \in \tau_X$, then $p(U) \in \tau_Y$. But this is obvious since $p^{-1}(p(U)) = U$ for injective maps.

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