On page 112 in the proof of Theroem 2.10 Hatcher defines the prism operators given by:
$P(\sigma)=\sum_i(-1)^iF\circ(\sigma\times Id_I)\mid[v_0,\cdots,v_i,w_i,\cdots,w_n]$
$F\circ(\sigma\times Id_I):\Delta^n\times I\rightarrow X \times I \rightarrow Y$, where $F$ is the homotopy from $f$ to $g$ and $\sigma$ is a singular simplex.
Is there an intuative way of thinking about what these operators do? Why is there an alternating sum? I've been tring to think about how an n-simplex in X that can be thought of as the bottom of $\Delta^n\times I$ can be mapped to an (n+1)-simplex in $\Delta^n\times I$ in any meaningful way. In essence I do not understand the link between the prism $\Delta^n\times I$ and the prim operators. Any simple example would be much appreciated.
Edit: In terms of a 1-simplex with prism $\Delta^1\times I$ we have
$P(\sigma)=F\circ(\sigma\times Id_I)\mid[v_0,w_0,w_1]-F\circ(\sigma\times Id_I)\mid[v_0,v_1,w_1]$
This maps a 1-simplex in $X$ to a 2-chain in $Y$ where the boundary is the "square" on the outside of $\Delta^1\times I$. So as I understand it we take some map from the bottom of $\Delta^1\times I$ to $X$ which is a 1-simplex and are then applying this map to a 2-simplex such as $[v_0,w_0,w_1]$. How can we apply a map $\sigma:\Delta^n\rightarrow X$ to $\Delta^{n+1}$?
Best Answer
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.
[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.