You don't want to prove that $\Delta \cap \Delta'$ is a simplex for each $\Delta \in K$ and $\Delta' \in L$, because it's false.
For a counterexample, take a regular hexagon in $\mathbb R^2$ with vertices $A,B,C,D,E,F$, let $K=\Delta$ be the 2-simplex with vertices $A,C,E$, and let $L=\Delta'$ be the 2-simplex with vertices $B,D,F$, and so $\Delta \cap \Delta'$ is a smaller regular hexagon. Of course that smaller hexagon is the realization of a simplicial complex, in fact any regular hexagon is the realization of a simplicial complex with one extra vertex in the center, and six 2-simplices altogether.
You should be able to find a simplicial complex structure on $|K| \cap |L|$ by carefully keeping this subdivision concept in mind. It'll be messy, if you want to do it with full rigor. But the key idea should be that the intersection of any simplex $\Delta$ of $K$ and any simplex $\Delta'$ of $L$ can be subdivided into a simplicial complex. Of course you have to do these subdivisions over all choices of $\Delta$ and $\Delta'$, to make sure that they all fit together into a simplicial complex structure on $|K| \cap |L|$.
Using the assumption that $K$ and $L$ are finite simplicial complexes, I think the best approach to writing down a rigorous proof is probably as a double induction. For the primary induction, you start with $L = \emptyset$ and then add one simplex of $L$ at a time, writing $L$ as the union of subcomplexes
$$L_1 \subset L_2 \subset \cdots \subset L_I
$$
where each $L_i$ is obtained from $L_{i-1}$ by adding a single simplex $\tau_i$. Assuming that $K$ and $L_{i-1}$ have been subdivided so that their intersection is a subcomplex of each, then you add $\tau_i$. Now you do a secondary induction on the dimension of the skeleta of $K$, writing $K$ as the union of its skeleta $K^{(0)} \subset K^{(1)} \subset \cdots \subset K^{(d)}=K$. You first subdivide $\tau_i$ at each point where its interior intersects $K^{(0)}$, after which $K^{(0)} \cap L_i$ will be a subcomplex of each of $K^{(0)}$ and $L_i$. In the induction step, assuming that $K^{(j)} \cap L_i$ is a subcomplex of each of $K^{(j)}$ and $L_i$, you then further subdivide $\tau_i$ so that $K^{(j+1)} \cap L_i$ is a subcomplex of each of $K^{(j+1)}$ and $L_i$.
Having looked at your edit, it seems like the confusion is around understanding how the prism operator is actually defined.
It seems like the notation $F\circ(\sigma\times \text{Id}_I)\mid[v_0,w_0,w_1]$ is the source of confusion. You've correctly identified that this map is a map from $\Delta^2$ to $Y$, but it seems like you're not completely clear on how this map is defined. So I'll explain.
The map $F\circ(\sigma\times \text{Id}_I)\mid[v_0,w_0,w_1]$ is a composition of the following three maps.
- The (unique) affine map from $\Delta^2$ to $\Delta^1 \times I$ that sends
$$(0, 0) \mapsto ((0), 0), \ \ \ \ \ (1, 0) \mapsto ((0), 1), \ \ \ \ \ (0, 1) \mapsto ((1), 1).$$
- The map from $\Delta^1 \times I$ to $X \times I$ that sends
$$ ((t_0), s) \mapsto (\sigma(t_0), s), \ \ \ \ \ \ \ \ \text{for all } (t_0) \in \Delta^1, \ s\in I,$$
where $\sigma : \Delta^1 \to X$ is the singular $1$-simplex that we started with.
- The map from $X \times I$ to $Y$ that sends
$$ (x, s) \mapsto F(x, s), \ \ \ \ \ \ \ \ \text{for all } x \in X, \ s\in I,$$
where $F : X \times I \to Y$ is the homotopy.
[An affine map is a composition of a linear transformation and a translation. An affine map from $\Delta^2$ to some other space is fully specified once you've specified the images of any three points in $\Delta^2$; in particular, it is fully specified once you've specified the images of the three corners of $\Delta^2$. In our example, our affine map maps the triangle $\Delta^2$ rigidly onto the triangle in $\Delta^1 \times I$ whose corners are situated at $((0), 0)$, $((0), 1)$ and $((1), 1)$.]
The map $F\circ(\sigma\times \text{Id}_I)\mid[v_0,v_1,w_1]$ is defined in a similar way. The only difference is that the affine map we use is the affine map that sends
$$(0, 0) \mapsto ((0), 0), \ \ \ \ \ (1, 0) \mapsto ((1), 0), \ \ \ \ \ (0, 1) \mapsto ((1), 1).$$
Edit: I thought I'd expand on this and describe how to prove that
$$ g_\# (\sigma) - f_\# (\sigma) = P (\partial \sigma) + \partial(P(\sigma))$$
when $\sigma$ is a singular $1$-simplex. I'll do so using my notation based on affine maps, rather than using Hatcher's notation.
This will help you see where the minus signs come in, as per your request below. And hopefully, it will help you follow the proof in general. (Yes, I also found the Hatcher's proof difficult to understand because of the notation - I didn't get my head around it until I rewrote everything in terms of affine maps.)
So let's get started. As already discussed, we start with a $1$-simplex $\sigma$ on $X$. Applying the prism operator $P$ on $\sigma$, we get $P(\sigma)$, which is a $2$-chain on $Y$:
$$ P(\sigma) = F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0,0)\mapsto ((0), 0) \\ (1, 0) \mapsto ((0), 1) \\ (0, 1) \mapsto ((1), 1)} - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0,0)\mapsto ((0), 0) \\ (1, 0) \mapsto ((1), 0) \\ (0, 1) \mapsto ((1), 1)}.$$
Applying the boundary operator $\partial$ to $P(\sigma)$, we get $\partial(P(\sigma))$, which is a $1$-chain on $Y$:
$$ \partial P(\sigma) = P(\sigma) \circ \text{Affine}_{(0)\mapsto (1, 0) \\ (1) \mapsto (0, 1)} - P(\sigma) \circ \text{Affine}_{(0)\mapsto (0, 0) \\ (1) \mapsto (0, 1)} + P(\sigma) \circ \text{Affine}_{(0)\mapsto (0, 0) \\ (1) \mapsto (1, 0)}.$$
Substituting in our expression for $P(\sigma)$ and expanding all the terms, we obtain the expression
\begin{align}
\partial P(\sigma) = & \ F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 1) \\ (1) \mapsto ((1), 1) } - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((1), 0) \\ (1) \mapsto ((1), 1) } \\ & \ - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 1) } + F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 1) }\\ & \ + F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((0), 1) } - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 0)}.
\end{align}
The two terms on the second line cancel each other out, leaving us with the expression
\begin{align}
\partial P(\sigma) = & \ F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 1) \\ (1) \mapsto ((1), 1) } - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((1), 0) \\ (1) \mapsto ((1), 1) } \\ & \ + F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((0), 1) } - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 0)} .
\end{align}
Meanwhile, if we apply the boundary operator $\partial$ to $\sigma$, then we get $\partial\sigma$, which is a $0$-chain on $X$:
$$ \partial\sigma = \sigma \circ \text{Affine}_{(.)\mapsto (1) } - \sigma \circ \text{Affine}_{(.)\mapsto (0) }. $$
Applying the prism operator $P$ to $\partial\sigma$, we get $P(\partial\sigma)$, which is a $1$-chain on $Y$:
$$ P(\partial\sigma) = F \circ (\partial\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((.), 0) \\ (1) \mapsto ((.), 1)}.$$
Substituting in our expression for $\partial\sigma$, we see that
\begin{align}
P(\partial\sigma) = F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((1), 0) \\ (1) \mapsto ((1), 1)} - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((0), 1)}.
\end{align}
Thus
\begin{align}
\partial P(\sigma) + P(\partial \sigma) = & \ F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 1) \\ (1) \mapsto ((1), 1) } - F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 0)} .
\end{align}
But $F(x, 1) = g(x)$ and $F(x, 0) = f(x)$ for all $x \in X$, since $F$ is a homotopy from $f$ to $g$. So
$$ F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 1) \\ (1) \mapsto ((1), 1) } = g \circ \sigma = g_\#(\sigma)$$
and
$$ F \circ (\sigma \times \text{Id}_I) \circ \text{Affine}_{(0)\mapsto ((0), 0) \\ (1) \mapsto ((1), 0)} = f \circ \sigma = f_\#(\sigma).$$
Thus
$$ \partial P(\sigma) + P(\partial \sigma) = g_\#(\sigma) - f_\#(\sigma),$$
which completes the proof.
Best Answer
Lemma. The intersection of two maximal simplices is a simplex, and it is the convex hull of its vertices.
I'm not going to prove this, and it could be that a careful proof would also solve the general problem you're asking. If I wanted to prove it, I might resort to coordinates: assume that the original simplex $\sigma$ is the standard $d$-simplex with its standard barycentric coordinates. Then each point in $\sigma \times I$ has coordinates, and the job would be to identify the points in $\sigma_{i}$, $\sigma_{j}$, and their intersection. Maybe we wouldn't need to resort to coordinates and could determine the intersection of $\sigma_i$ and $\sigma_j$ from their vertices.
Anyway, assume that the lemma is true and suppose that $F_{i}$ and $F_{j}$ are faces of $\sigma_{i}$ and $\sigma_{j}$, respectively, and let $F = \sigma_{i} \cap \sigma_{j}$. Then $F_{i} \cap F_{j} \subseteq F$, and rather than considering $F_{i} \cap F_{j}$, we can instead let $F_{i}' = F_{i} \cap F$ and $F_{j}' = F_{j} \cap F$, and consider $F_{i} \cap F_{j} = F_{i}' \cap F_{j}'$. Now $F_{i}'$ and $F_{j}'$ are both contained in $F$, and so all of the involved vertices are in general position, so $F_{i} \cap F_{j} = F_{i}' \cap F_{j}'$ is the convex hull of the vertices in the intersection.