Is possible to generalise openness of projections and embeddability of factors of a product space to an initial topology

Question

Let be $\frak F$ a collection of functions defined into a set $X$ so that for any $f\in\cal F$ let be $\cal T_f$ a topology for the image of $Y_f$ of $f$: so it is a well knew result that then the collection
$$
\mathcal S:=\big\{f^{-1}[A]:(f,A)\in\mathfrak F\times \mathcal T_f\big\}
$$
is a subbase for a topology $\cal T_\frak F$ which is called initial topology corresponding to $\frak F$.

So, I know that when $X$ is a product and so when $\frak F$ is just the collection of projections then these are open with respect $\cal T_\frak F$; moreover I know that if $X$ is a product then $Y_f$ for any $f\in\frak F$ is embeddable into $X$ when this is endowed with $\cal T_\frak F$: so, I asked to me if this is true even when $X$ is not a product or when $\frak F$ is another collection but unfortunately I was not able to say anything about so that I thought to put a specific question where I ask clarification. So could someone help me, please?

Answer 1

Analysing the corresponding proofs for product topology I elaborated by myself a possibile answer: what I did seemed to me reasonable to the point that I deliberated to condivide it as wiki answer, hoping someone can add some details or correct it wheter it is not correct or wheter it has some imperfections. Thanks in advance for read it!!!

First of all, we graphically represent the domain and the codomain of any function $f$ with the symbols $D_f$ and $C_f$ and in particular if $\frak F$ is a collection of function with same domain or same codomain the we omit the subscript; additionally, we represent the final topology generated by $\frak F$ with the symbol $\cal T_\frak F$ and in particular if $\frak F$ is a singleton the we write simply $\cal T_f$ where $f$ is the unique element of $\frak F$.

So, let's we put $$ f(x):=\begin{cases}-1,\,\text{if $x<0$}\\0,\,\text{if $x=0$}\\1,\,\text{if $x>0$}\end{cases} $$ for any $x\in\Bbb R$ so that the equality $$ \tag{0}\label{0}\{0\}=f^{-1}\big[(-1,1)\big] $$ holds: thus by \eqref{0} we conlcude that $\{0\}$ is open in $\Bbb R$ with respect $\cal T_f$ but $f[0]$ is just $\{0\}$ which is not surely open in $\Bbb R$ with respect the usual topology. So we conclude that the elements of $\frak F$ can be not open when $X$ is endowed with the corresponding initial topology so that the statement is not generally true but it can actually be when for each $(\boldsymbol f,\boldsymbol S)\in\prod_{i\in n}\mathfrak F\times\prod_{i\in n}\big(\mathscr P(C_{\boldsymbol f(i)})\setminus\{\emptyset\}\big)$ with $n\in\omega_+$ the intersection $$ I(\boldsymbol f,\boldsymbol S):=\bigcap_{i\in n}\big(\boldsymbol f(i)\big)^{-1}[\boldsymbol S(i)] $$ is not empty as to follow it is proved.

First of all, for practical reasons let's we give the following definition

Definition 0

A set $D$ is said finitely generated by the collection of functions $\frak F$ there defined when for each $(\boldsymbol f,\boldsymbol S)\in\prod_{i\in n}\mathfrak F\times\prod_{i\in n}\big(\mathscr P(C_{\boldsymbol f(i)})\setminus\{\emptyset\}\big)$ with $n\in\omega_+$ the intersection $$ I(\boldsymbol f,\boldsymbol S):=\bigcap_{i\in n}\big(\boldsymbol f(i)\big)^{-1}[\boldsymbol S(i)] $$ is not empty.

So, let's we prove the following lemma

Lemma 1

If $D$ is finitely generated by the collection of functions $\frak F$ then then any $f\in\frak F$ is surjective onto its codomain and additionally if $(\boldsymbol f,\boldsymbol S)\in\prod_{i\in n}\mathfrak F\times\prod_{i\in n}\big(\mathscr P(C_{\boldsymbol f(i)})\setminus\{\emptyset\}\big)$ then for any $f\in\frak F$ the equality $$ f\big[I(\boldsymbol f,\boldsymbol S)\big]= \begin{cases} \boldsymbol S(\iota),\,\text{ if $f=\boldsymbol f(\iota)$ for any $\iota\in n$}\\ C_f,\,\text{otherwise} \end{cases} $$ holds.

Proof. So if $D$ is finitely generated by $\frak F$ then for any $c_f\in C_f$ with $f\in\frak F$ the set $$ I(f,c_f)\equiv f^{-1}[c_f] $$ is not empty so that $f$ is surjective. Now let be $\boldsymbol\varphi\in\prod_{i\in n}\frak F$ defined as $$ \boldsymbol\varphi(i):=\boldsymbol f(i) $$ for each $i\in n$ and so for any $s_\iota\in\boldsymbol S(\iota)$ with $\iota\in n$ let be $\boldsymbol\Phi\in\prod_{i\in n}\big(\mathscr P(C_{\boldsymbol\varphi(i)}\setminus\{\emptyset\}\big)$ defined as $$ \boldsymbol\Phi(i):=\begin{cases} \{s_\iota\},\,\text{if $i=\iota$}\\ \boldsymbol S(i),\,\text{otherwise} \end{cases} $$ for each $i\in n$: therefore, we observe that the inclusion $$ I(\boldsymbol\varphi,\boldsymbol\Phi)\subseteq I(\boldsymbol f,\boldsymbol S) $$ holds so that there exists $d_\iota\in I(\boldsymbol\varphi,\boldsymbol\Phi)$ lying into $\big(\boldsymbol f(\iota)\big)^{-1}[s_\iota]$ or rather such that $$ [\boldsymbol f(\iota)](d_\iota)=s_\iota $$ which proves that $$ \big[\boldsymbol f(\iota)\big][I(\boldsymbol f,\boldsymbol S)]=S_\iota $$ Now let be $c_f\in C_f$ with $f\in\frak F$ such that $f$ is not equal to $\boldsymbol f(i)$ for each $i\in n$ and thus let be now $\boldsymbol\psi\in\prod_{i\in S(n)}\frak F$ defined as $$ \boldsymbol\psi(i):=\begin{cases} \boldsymbol f(i),\,\text{if $i<n$}\\ f,\,\text{if $i=n$} \end{cases} $$ for each $i\in S(n)$ and thus let be $\boldsymbol\Psi\in\prod_{i\in S(n)}\mathcal T_{\boldsymbol\psi(i)}$ defined as $$ \boldsymbol\Psi(i):=\begin{cases} \boldsymbol S(i),\,\text{if $i<n$}\\ \{c_f\},\,\text{if $i=n$} \end{cases} $$ for each $i\in S(n)$ so that the inclusion $$ I(\boldsymbol\psi,\boldsymbol\Psi)\subseteq I(\boldsymbol f,\boldsymbol S) $$ holds and so by the same arguments above applied we conclude that $$ f[I(\boldsymbol f,\boldsymbol S)]=C_f $$

Finally let's we prove that if each $C_f$ with $f\in\frak F$ is equipped with a topology $\cal T_f$ then any $f\in\frak F$ is an open function when $D$ is endowed with respect the initial topology $\cal T_\frak F$ corresponding to $\frak F$, provided it is finitely generated with respect $\frak F$

Theorem 2

Let be $\cal T_f$ with $f\in\frak F$ a topology for $C_f$: so if $D$ is finitely generated with respect $\frak F$ then any $f\in\frak F$ is an open function when $D$ is equipped with initial topology $\cal T_\frak F$ corresponding to $\frak F$.

Proof. So if $d\in O$ with $O\in\cal T_\frak F$ then $$ d\in I(\boldsymbol f,\boldsymbol O)\subseteq O $$ for any $(\boldsymbol f,\boldsymbol O)\in\prod_{i\in n}\mathfrak F\times\prod_{i\in n}\mathcal T_{\boldsymbol f(i)}$. So each $f\in\frak F$ is such that $$ f(d)\in f[I(\boldsymbol f,\boldsymbol O)]\subseteq f[O] $$ but by the last lemma we know that $f[I(\boldsymbol f,\boldsymbol O)]$ is just $\boldsymbol O(i)$ for any $i\in n$ or $C_f$ which are both open so that we finally conclude $f[O]$ is open.

So, for check, we observe that the product $\prod\frak X$ of any collection $\frak X$ is finitely generated with respect the projections collection: in fact, if $Y_i$ for $i\in n$ with $n\in\omega_+$ is a not empty subset of $X_i\in\frak X$ then for any $i\in n$ we can pick $y_i\in Y_i$ so that fixed $x\in\prod\frak X$ the position $$ y(X):=\begin{cases} y_i\,\text{if $X=X_i$ for any $i\in n$}\\ x(X)\,\text{otherwise} \end{cases} $$ with $X\in\frak X$ defines an $y\in\prod\frak X$ lying into $\bigcap_{i\in n}\pi^{-1}_{X_i}[Y_i]$. So by theorem $2$ we conclude that any projection of $\prod\frak X$ is open with respect Tychonoff topology, as after all it is very knew. Moreover, we observe that the lemma $1$ is just a well knew result when $D$ is just a product and so when $\frak F$ is just the corresponding projections collection.

Now we observe that the subspace topology of indiscrete topology is just the indiscrete topology so that let's we observe that the initial topology on $\Bbb R$ corresponding to any constant map is the indiscrete and thus $\Bbb R$ endowed with the euclidean topology is not embeddable in $\Bbb R$ endowed with the initial topology corresponding to any constant map because any indiscrete space is hausdorff: so we conclude that any $C_f$ with $f\in\frak F$ is not embeddable into $X$ when it is endowed with the corresponding initial topology so that the statement is not generally true but it can actually be when for any $f_0\in\frak F$ there exists $\boldsymbol c_{f_0}\in\prod_{\mathfrak F\setminus\{f_0\}}C_f$ such that $$ V(c_{f_0}):=f^{-1}_0[c_{f_0}]\cap\Biggl(\bigcap_{f\in\mathfrak F\setminus\{f_0\}}f^{-1}[\boldsymbol c_{f_0}(f)]\Biggl) $$ is not empty for each $c_{f_0}\in C_{f_0}$, as to follow is proved.

So, first of all for convenience we give the following definition.

Definition 3

A set $D$ is a box with respect the collection of functions $\frak F$ there defined if for any $f_0\in\frak F$ there exists $\boldsymbol c_{f_0}\in\prod_{\mathfrak F\setminus\{f_0\}}C_f$ such that $$ V(c_{f_0}):=f^{-1}_0[c_{f_0}]\cap\Biggl(\bigcap_{f\in\mathfrak F\setminus\{f_0\}}f^{-1}[\boldsymbol c_{f_0}(f)]\Biggl) $$ is not empty for each $c_{f_0}\in C_{f_0}$.

So, let's we prove the following theorem.

Theorem 4

If a set $D$ is a box with respect the collection of functions $\frak F$ and thus if each $C_f$ with $f\in\frak F$ is equipped with a topology $\cal T_f$ then it is embeddable into $X$ when this is endowed with the initial topology $\cal T_\frak F$ corresponding to $\frak F$.

Proof. So, if $D$ is a box with respect the collection of functions $\frak F$ then fixed $f_0\in\frak F$ for any $c_{f_0}\in C_{f_0}$ we can put $$ d(C_{f_0}):=\sigma\big(V(c_{f_0})\big) $$ where $\sigma$ is a choice function so that let be $g_0$ the function $$ g_0:C_{f_0}\ni c_{f_0}\longrightarrow d(c_{f_0})\in D $$ So, we observe that $$ g_0(c_{f_0})\in V(c_{f_0})\subseteq f^{-1}_0[c_{f_0}] $$ for any $c_{f_0}\in C_{f_0}$ so that the equality $$ \tag{1}\label{1}f_0\circ g_0=\operatorname{id}_{C_{f_0}} $$ holds; after all, if $d_{f_0}\in g_0[C_{f_0}]$ then there exists $c_{f_0}\in C_{f_0}$ such that $$ d_{f_0}=g_0(c_{f_0})=d(c_{f_0})\in V\big(c_{f_0}\big)\subseteq f^{-1}_0[c_{f_0}] $$ so that $f_0(d{f_0})$ is $c_{f_0}$ and thus $$ g_0\big(f_0(d_{f_0})\big)=g_0(c_{f_0})=d(c_{f_0})=d{f_0} $$ which proves that the equality $$ \tag{2}\label{2}g_0\circ f_0=\operatorname{id}_{g_0[C_{f_0}]} $$ holds: so by \eqref{1} and by \eqref{2} we conclude that $f_0$ and $g_0$ are bijection which are the inverse of each other. So given $f\in\frak F$ we observe that for each $c_{f_0}\in C_{f_0}$ the equality $$ \big(f\circ g_0\big)(c_{f_0})=\begin{cases} c_{f_0}\,\text{if $f=f_0$}\\ \boldsymbol c_{f_0}(f)\,\text{otherwise} \end{cases} $$ holds so that $f\circ g_0$ is the identity on $C_{f_0}$ or a constant map which are both continuous and thus by the universal mapping theorem for initial topologies we conclude that $g_0$ is continuous; after all, even $f_0$ is continuous by definition of initial topology: so, we conclude that $g_0$ is an embedding.

Finally, for check, we observe that if $\frak X$ is a collection of topological spaces then fixed $\xi_0\in\prod\frak X$ for any $x_0\in X_0$ with $X_0\in\frak X$ we can put $$ \xi(X):=\begin{cases} x_0,\,\text{if $X=X_0$}\\ \xi_0(X),\,\text{otherwise} \end{cases} $$ with $X\in\frak X$ defining $\xi\in\prod\frak X$ lying into $$ V(x_0)\equiv\pi^{-1}_{X_0}[x_0]\cap\Biggl(\bigcap_{X\in\mathfrak X\setminus\{X_0\}}\pi^{-1}_X\big[\xi_0[X]\big]\Biggl) $$ which proves that any product is a box with respect the corresponding projection functions.