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This answer slot available for anyone to use.

For your concern about Kodaira embedding theorem:

This follows from the inverse function theorem and is completely local (being an analytic subvariety is an local condition).

Let $k\le n$, $U\subset \mathbb C^k$ be open and $f : U \to \mathbb C^n$ be a holomorphic map so that it is immersed at $p\in U$. By composing with a complex linear map $\mathbb C^n \to \mathbb C^n$, we assume $$ \operatorname{Im} df(p) = \mathbb C^k \times \{0\} \subset \mathbb C^n.$$

Then the function $$ F: U\times \mathbb C^{n-k} \to \mathbb C^n, \ \ \ F(z, u) = f(z) + (0,u)$$

has invertible $dF (p, 0)$. Thus there is $V\subset U \times \mathbb C^{n-k}$, W\subset \mathbb C^n$ so that $F : V \to W$ is a biholomorphism, and in particular,

\begin{align} \operatorname{Im} f (U\cap V) &= \{ F(z, u) : u_1 = \cdots u_{n-k} = 0\}\\ &= \{ w \in W : F^{-1}\circ u_1 = \cdots =F^{-1}\circ u_{n-k} = 0\}. \end{align}

and $F^{-1} \circ u_i$ are holomorphic functions on $W$. Thus the image is an analytic subvariety.

If your concern is why

Solve alphanumeric puzzle $(HE) \times (EH) = (WHEW).$

This is a self-answer query. I came across a recent mathSE query that presented an interesting problem, and took some time solving it. When I looked for the query to respond, I couldn't find it. I suspect that the problem may have been closed or deleted.

The real challenge is to avoid brute force and to present a completely analytical solution. That is why I think it is worthwhile to present the problem and solution.

Problem
In the multiplication problem below, the letter $H$ is used to symbolize the same digit (one of the values $0$ through $9$) throughout the problem.
Similarly, the letter $E$ symbolizes a specific digit throughout the problem, and the letter $W$ symbolizes a specific digit throughout the problem.

The letters $H,E,W$ symbolize different digits from each other. No number is allowed to have its leftmost digit $= 0$. Therefore, you know immediately that none of $H,E,$ or $W$ can equal $0$.

\begin{array}{ r r r r} & & H & E \\ \times & & E & H \\ \hline W & H & E & W \\ \end{array}

Determine the values for $H,E,$ and $W$.

======

Answer:

! Since $H \times E \equiv W \pmod{10},$ and $H,E,W$ are all distinct, you know that neither $H$ nor $E$ can equal $1$. Similarly, since $W \neq 0$, you know that neither $H$ nor $E$ can equal $5$. Therefore, you also know that $W \neq 5$.
!
! So, at this point, you know that $H,E \in \{2,3,4,6,7,8,9\}$ and
! that $W \in \{1,2,3,4,6,7,8,9\}.$
! ! You also know that $W = 9$ is impossible, because that would prevent either $H$ or $E$ from equaling $9$. Then, the largest possible product would be $(87 \times 78) < 9000$, which would violate the constraint that the leftmost digit of the product is $W = 9$.
!
! Similarly, $W$ can't equal $8$, because $97 \times 79 < 8000.$ ! ! So, at this point, you know that $W \in \{1,2,3,4,6,7\}.$
! ! If $W = 3$, then $H \times E \equiv W = 3\pmod{10}$.
! Therefore, $H$ and $E$ must both be \in {3,7,9}.$ >! The only possibility here, $H,E$ = 7,9$, in some order, is impossible because $(79) \times (97) > 3999.$ !
! So, at this point, you know that $W \in \{1,2,4,6,7\}.$ !
! $W = 1$ is impossible, because that would require $H,E = 7,3$,
! some order, and $(37) \times (73) > 1999.$ ! ! So, at this point, you know that $W \in \{2,4,6,7\}.$ ! ! $W = 7$ is impossible, because that would require $H,E = 9,3$,
! in some order, and $(39) \times (93) < 7000.$ ! ! So, at this point, you know that $W \in \{2,4,6\}.$ ! ! Suppose that $(10)H + E$ \equiv k \pmod{9} ~:~ k \in {1,2,\cdots,9}.$ >! By [Casting out 9's](https://en.wikipedia.org/wiki/Casting_out_nines) you know that $(10)E + H$ and $(E + H)$ are each also $\equiv k\pmod{9}$. >! >! This implies that $(2W) + H + E \equiv k^2 \pmod{9} \implies $ >! $2W \equiv (k^2 - k) = k(k-1) \pmod{9}.$ >! >! As $k$ ranges from $1$ through $9$, $k(K-1)$ is never equal to either $8$ or $4$ \pmod{9}.$ ! ! From this you know that $w$ can't equal either $2$ or $4$, so $w = 6$
! and $H,E \in \{2,3,4,7,8,9\}$. ! ! Further, since $59 times 95 < 6000$, you know that
! $H,E \in \{7,8,9\}.$ ! ! Therefore, $H,E$ = $7,8$, in some order.
! Since $78 \times 87 = 6786$, you know that $H,E = 7,8.$ respectively.

This answer slot available for anyone to use.

! Since $H \times E \equiv W \pmod{10}, and $H,E,W$ are all distinct, you know that neither $H$ nor $E$ can equal $1$. Similarly, since $W \neq 0$, you know that neither $H$ nor $E$ can equal $5$. Therefore, you also know that $W \neq 5$. !> !> So, at this point, you know that $H,E \in {2,3,4,6,7,8,9}$ and !> that $W \in {1,2,3,4,6,7,8,9}.$ !> !> You also know that $W = 9$ is impossible, because that would prevent either $H$ or $E$ from equaling $9$. Then, the largest possible product would be $(87 \times 78) < 9000$, which would violate the constraint that the leftmost digit of the product is $W = 9$. !> !> Similarly, $W$ can't equal $8$, because $97 \times 79 < 8000.$ !> !> So, at this point, you know that $W \in {1,2,3,4,6,7}.$ !> !> If $W = 3$, then $H \times E \equiv W = 3\pmod{10}$. !> Therefore, $H$ and $E$ must both be \in \{3,7,9\}.$
!> The only possibility here, $H,E$ = 7,9$, in some order, is impossible because $(79) \times (97) > 3999.$ !> !> So, at this point, you know that $W \in {1,2,4,6,7}.$ !> !> $W = 1$ is impossible, because that would require $H,E = 7,3$, !> some order, and $(37) \times (73) > 1999.$ !> !> So, at this point, you know that $W \in {2,4,6,7}.$ !> !> $W = 7$ is impossible, because that would require $H,E = 9,3$, !> in some order, and $(39) \times (93) < 7000.$ !> !> So, at this point, you know that $W \in {2,4,6}.$ !> !> Suppose that $(10)H + E$ \equiv k \pmod{9} ~:~ k \in \{1,2,\cdots,9\}.$ By Casting out 9's you know that $(10)E + H$ and $(E + H)$ are each also $\equiv k\pmod{9}$. !> !> This implies that $(2W) + H + E \equiv k^2 \pmod{9} \implies $
$2W \equiv (k^2 - k) = k(k-1) \pmod{9}.$ !> !> As $k$ ranges from $1$ through $9$, $k(K-1)$ is never equal to either $8$ or $4$ \pmod{9}.$ !> !> From this you know that $w$ can't equal either $2$ or $4$, so $w = 6$ <br> and $H,E \in {2,3,4,7,8,9}$. !> !> Further, since $59 times 95 < 6000$, you know that !> $H,E \in {7,8,9}.$ !> !> Therefore, $H,E$ = $7,8$, in some order. !> Since $78 \times 87 = 6786$, you know that $H,E = 7,8.$ respectively.

! Since $H \times E \equiv W \pmod{10},$ and $H,E,W$ are all distinct, you know that neither $H$ nor $E$ can equal $1$. Similarly, since $W \neq 0$, you know that neither $H$ nor $E$ can equal $5$. Therefore, you also know that $W \neq 5$.
!
! So, at this point, you know that $H,E \in \{2,3,4,6,7,8,9\}$ and
! that $W \in \{1,2,3,4,6,7,8,9\}.$
! ! You also know that $W = 9$ is impossible, because that would prevent either $H$ or $E$ from equaling $9$. Then, the largest possible product would be $(87 \times 78) < 9000$, which would violate the constraint that the leftmost digit of the product is $W = 9$.
!
! Similarly, $W$ can't equal $8$, because $97 \times 79 < 8000.$ ! ! So, at this point, you know that $W \in \{1,2,3,4,6,7\}.$
! ! If $W = 3$, then $H \times E \equiv W = 3\pmod{10}$.
! Therefore, $H$ and $E$ must both be \in {3,7,9}.$ >! The only possibility here, $H,E$ = 7,9$, in some order, is impossible because $(79) \times (97) > 3999.$ !
! So, at this point, you know that $W \in \{1,2,4,6,7\}.$ !
! $W = 1$ is impossible, because that would require $H,E = 7,3$,
! some order, and $(37) \times (73) > 1999.$ ! ! So, at this point, you know that $W \in \{2,4,6,7\}.$ ! ! $W = 7$ is impossible, because that would require $H,E = 9,3$,
! in some order, and $(39) \times (93) < 7000.$ ! ! So, at this point, you know that $W \in \{2,4,6\}.$ ! ! Suppose that $(10)H + E$ \equiv k \pmod{9} ~:~ k \in {1,2,\cdots,9}.$ >! By [Casting out 9's](https://en.wikipedia.org/wiki/Casting_out_nines) you know that $(10)E + H$ and $(E + H)$ are each also $\equiv k\pmod{9}$. >! >! This implies that $(2W) + H + E \equiv k^2 \pmod{9} \implies $ >! $2W \equiv (k^2 - k) = k(k-1) \pmod{9}.$ >! >! As $k$ ranges from $1$ through $9$, $k(K-1)$ is never equal to either $8$ or $4$ \pmod{9}.$ ! ! From this you know that $w$ can't equal either $2$ or $4$, so $w = 6$
! and $H,E \in \{2,3,4,7,8,9\}$. ! ! Further, since $59 times 95 < 6000$, you know that
! $H,E \in \{7,8,9\}.$ ! ! Therefore, $H,E$ = $7,8$, in some order.
! Since $78 \times 87 = 6786$, you know that $H,E = 7,8.$ respectively.