boundary( geometric realization of the standard n-simplex) is not equal to the geometric realization of the boundary(standard n-simplex) in general

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Consider $|\Delta^n|$ the geometric realization of the standard n-simplex. I know that the $|\delta \Delta^n|=\delta|\Delta^n|$ isn't true in general, whereby $\delta \Delta^n$ is the boundary of the n-simplex. If i consider $\delta_{\mathbb{R}^{n+1}}|\Delta^n|$, the dimension of $\delta_{\mathbb{R}^{n+1}}|\Delta^n|$ is n but i want to embed the boundary of $|\Delta^n|$ in a subset of $\mathbb{R}^{n+1}$, such that the equality $|\delta \Delta^n|=\delta|\Delta^n|$ will be true; maybe $\delta_{U}|\Delta^n|$ with $U=\{x\in \mathbb{R}^{n+1}:\sum_{k=0}^nx_i=1\}$ with $dim_\mathbb{R}U=n$, is it possible? How does the set $\delta_{U}|\Delta^n|$look like? Is it the sphere $S^{n+1}$ or only homeomorphic?
I hope you understand my problem. Regards

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I'm confused by your question, but I'll try to answer it:

$|\Delta^n | = \{ x \in \mathbb{R}^{n+1} : \sum x_i = 1 \}$

We call $(1,0,0....)$ the first vertex, $(0,1,0....)$ the second, etc. By the $i$th face, I will mean the face opposite to the $i$th vertex.

The $i$th face of the standard $n$-simplex, $d_i |\Delta^n| = \{x \in |\Delta^n| : x_i = 0 \}$.

The boundary $\partial |\Delta^n|$ is the union of all the faces $0 \leq i \leq n$.

Each face is a copy of $|\Delta^{n-1}|$. For example the map

$(x_1, x_2) \to (x_1,0,x_2)$

gives a homeomorphism $|\Delta^{1}| \to d_1|\Delta^{2}|$.

Nothing simplicial will "look like" a sphere. The sphere is "smooth" and simplicial things have edges and corners (piecewise linear). But for the purposes of topology, $\partial|\Delta^{n+1}|$ and $S^n$ are the same, ie. homeomorphic.

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The problem is one of dimension. If you embed the $n$-simplex in $\mathbb R^{n+1}$ or higher, the geometric boundary will actually be the whole simplex! (Think about what happens when you embed an interval in $\mathbb R^2$.) So you need to embed the $n$-simplex in $\mathbb R^n$ for the boundaries to behave like you want them to.