Corollary 7.15 of Hartshorne states: If $f:Y\to X$ is a morphism of Noetherian schemes and $\mathcal{I}$ is a coherent sheaf of ideals on $X$, then there is a diagram $$ \begin{align*} &\overline{Y}&& \to &&\overline{X}\\ &\downarrow&&&&\downarrow\\ &Y&&\to&&X \end{align*} $$ where $\overline{X}$ is the blowup of $X$ along $\mathcal{I}$ and $\overline{Y}$ is the blowup of $Y$ along $f^{-1}\mathcal{I}\cdot \mathcal{O}_Y$.
I would like to apply this to a complex variety $V$ as follows: given two integers $1\leq n\leq m$, consider the map on product spaces defined by $$f:V^n\to V^m, (a_1,\dots,a_n)\to (a_1,\dots,a_n,a_n,\dots,a_n).$$ Now if $n\geq 2$ it seems that restricting the ideal sheaf of the small diagonal $\Delta_m\subset V^m$ gives me the ideal sheaf of the small diagonal $\Delta_n\subset V^n$, and hence we have a corresponding map
$$ \begin{align*} &\text{Bl}_{\Delta_n}V^n&& \to &&\text{Bl}_{\Delta_m}V^m\\ &\downarrow&&&&\downarrow\\ &V^n&&\to&&V^m \end{align*} $$ But if $n=1$, I am slightly confused as to what should go into the top left corner. It seems like there should be a copy of $V$ blown up along the zero ideal, but I don't know how to make sense of that. Given that the blowup should be a copy of $\text{Proj}(\oplus I^n)$, I believe there should be a copy of $V$ in the top left corner. But then I am confused with what the map $V\to \text{Bl}_{\Delta_m}V^m$ would be; since the diagram commutes I would expect $V$ to map into the exceptional divisor of $\text{Bl}_{\Delta_m}V^m$ but this should require the tangent bundle of $V$ to have a global section, which shouldn't always hold.
Could anyone shed some light as to if this reasoning is correct, and if so, what is the map $V\to \text{Bl}_{\Delta_m}V^m$? Thanks in advance.