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I think what is going on is the following.

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

Your theorem suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. Define a different map $V:F/\ker M \rightarrow \im M^{1/2}$ by $$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$ Again, this is well-defined. As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so $$ (Mx|My)_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)_0 = (PM^{1/2}x|PM^{1/2}y) = (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$ Thus $$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$ is an isometry, as you want.

I think what is going on is the following.

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

Your theorem suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. Define a different map $V:F/\ker M \rightarrow \im M^{1/2}$ by $$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$ Again, this is well-defined. As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so $$ (Mx|My)_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)_0 = (PM^{1/2}x|PM^{1/2}y) = (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$ Thus $$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$ is an isometry, as you want. Notice that $V$ is not (in general) onto. Finally, we can extend $V$ to the completion of $F/\ker M$, and then we will obtain a unitary transformation onto $\overline{\im M^{1/2}}$, completion with respect to $(\cdot|\cdot)_0$.

I think what is going on is the following. The confusion comes (as is often the case) from confusing "equals in a natural way" with "isomorphic".

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

Your "theorem" suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. I don't believe that this is correct, as if we use $U$ to move $(\cdot|\cdot)_0$ to $F/\ker M$ we get the new inner-product $$ \ip{\overline x}{\overline y}_0 := (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ which is exactly what I considered, and dismissed, above.

I think what is going on is the following.

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

Your theorem suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. Define a different map $V:F/\ker M \rightarrow \im M^{1/2}$ by $$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$ Again, this is well-defined. As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so $$ (Mx|My)_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)_0 = (PM^{1/2}x|PM^{1/2}y) = (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$ Thus $$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$ is an isometry, as you want.

I think what is going on is the following. The confusion comes (as is often the case) from confusing "equals in a natural way" with "isomorphic".

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U : \overline{\im M} \rightarrow F/\ker M$ as follows. Firstly notice that $\overline{\im M} = \overline{\im M^{1/2}}$ because $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$. Then define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. We have only define $U$ on $\im M^{1/2}$ but as $U$ is an isometry it extends to $\overline{\im M^{1/2}}$. Clearly $U$ is onto, and so $U$ is unitary, thus establishing the isomorphism between $\overline{\im M}$ and $F/\ker M$.

I think what is going on is the following. The confusion comes (as is often the case) from confusing "equals in a natural way" with "isomorphic".

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$. Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$ Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$. It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined. Notice that we have not established that $F/\ker M$ is complete for this inner-product.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$. Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$. This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$. By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

However this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete. The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$. Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$. This is not the inner product you want.

Instead let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows. Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$. This is well-defined, for if also $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$. Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$, $$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y) = (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$ Hence $U$ is an isometry. Clearly $U$ is surjective. So $$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im(M^{1/2}), (\cdot|\cdot) \big) $$ is an isometric linear isomorphism. This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$. In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the completion of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

Your "theorem" suggests that we give $\im(M^{1/2})$ a different inner-product, $$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$. I don't believe that this is correct, as if we use $U$ to move $(\cdot|\cdot)_0$ to $F/\ker M$ we get the new inner-product $$ \ip{\overline x}{\overline y}_0 := (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$ which is exactly what I considered, and dismissed, above.