I want to verify the following claim:
Any singular $n-$boundary "encloses" a homotopy of singular $n-$simplices (for $n \ge 1$).
Specifically, let $p \in \Delta_n$ be given, i.e. $p=(p_0,p_1,\dots,p_n) \in \mathbb{R}^{n+1}$, $\sum_{k=0}^n p_k =1$.
Let $\phi: \Delta_{n+1} \to X$ be continuous (this is called a singular $(n+1)-$simplex).
As I argue below, I think the following assertion "proves" the above heuristic claim -- I am pretty sure that the correctness of the assertion follows immediately from the definitions, but I would like someone to double-check this, since I have not seen it written anywhere else before:
Assertion: Between the two singular $n-$simplices, $$(\partial_i\phi)(r_0, \dots, r_n) = \phi(r_0, \dots, r_{i-1},0,r_i, r_{i+1}, \dots, r_n), \\ (\partial_j\phi)(s_0, \dots, s_n)=\phi(s_0,\dots,s_{j-1},0,s_j,s_{j+1},\dots, s_n),$$ called the $i$th and $j$th faces of the singular $(n+1)-$simplex $\phi$, respectively, the following is a homotopy, i.e. a continuous function $H: \Delta_n \times[0,1] \to X$ such that for all $p \in \Delta_n$, $H(p,0)=(\partial_i \phi)(p)$ and $H(p,1)=(\partial_j \phi)(p)$: $$ H(p,\!t)\!=\!\begin{cases} \phi(p_0,\!\dots\!, p_{i-1}, tp_i, (1\!-\!t)p_i + tp_{i+1},\!\dots\!, (1\!-\!t)p_{j-2}+tp_{j-1}, (1\!-\!t)p_{j-1}, p_j,\!\dots\!, p_n) & i < j \\ \phi(p_0,\!\dots\!, p_{j-1}, tp_j, (1\!-\!t)p_j + tp_{j+1},\!\dots\!, (1\!-\!t)p_{i-2}+tp_{i-1}, (1\!-\!t)p_{i-1}, p_i,\!\dots\!, p_n) & j<i \end{cases}$$ Additionally, it is also the case that $H(\Delta_n \times [0,1]) \subseteq \phi(\Delta_{n+1})$.
Unnecessarily Verbose Background Information
Note: singular $n-$simplices are continuous maps $\sigma: \Delta_n \to X$, so given two singular $n-$simplices $\sigma_1, \sigma_2$, a homotopy between them is a continuous function $H: \Delta_n \times [0,1] \to X$ such that, for $p \in \Delta_n$, $H(p,0)=\sigma_1(p)$ and $H(p,1)=\sigma_2(p)$.
What I mean by the image of a singular $m-$chain $0\not=\mathfrak{s}\in C_m(X)$: for any singular $m-$chain $\mathfrak{s}\not=0$ we have a decomposition as follows: $$ \mathfrak{s} = a_1\mathfrak{s}_1 + \dots + a_{\ell}\mathfrak{s}_{\ell} $$ where all of the $a_i \not=0$ and all of the $\mathfrak{s}_i$ are singular $m-$simplices, continuous maps $\mathfrak{s}_i: \Delta_m \to X$. Then we define the image of $\mathfrak{s}$ to be $$\bigcup_{j=1}^{\ell} \mathfrak{s}_j(\Delta_m) \subseteq X$$ i.e. the image of the "sum" of generators is just their union, and non-zero multiplicity does not affect the set of points we determine to be their image. For the singular $m-$chain $0$ (the additive identity of the free abelian group $C_m(X)$), we declare its image to be the empty set $\emptyset$.
When I say that a singular $n-$boundary $\partial(c)$ encloses a homotopy $\tilde{\phi}$ singular $n-$simplices, I mean that the image of the homotopy $\tilde{\phi}(\Delta_n \times [0,1]) \subseteq X$ in X is a subset of the image of the singular $(n+1)-$chain $c$ as defined above.
Attempt: Define the $(n+1)$th boundary map $\partial: C_{n+1}(X) \to C_n(X)$ by $$\partial:= \sum_{i=0}^{n+1} (-1)^i \partial_i \quad \text{(with the $i$th face $\partial_i$ to be defined later).}$$ (This is the definition used in Vick's Homology Theory -- note that $C_{n+1}(X)$ is the group of singular $(n+1)-$chains, i.e. the free abelian group whose generators are singular $(n+1)-$simplices). Given any $n-$boundary, by definition it equals $\partial(c)$ for some $a_1 c_1 + \dots + a_kc_k \in C_{n+1}(X)$, where all of the $c_i$ are singular $(n+1)-$simplices. Since $\partial$ is a group homomorphism, we have that $$\partial(c)=\partial(a_1c_1 + \dots + a_k c_k)=a_1\partial(c_1) + \dots + a_k\partial(c_k), $$ so without loss of generality we consider only the case of $n-$boundaries of the form $\partial(\phi)$ for $\sigma$ a singular $(n+1)-$simplex, $\phi: \Delta_{n+1} \to X$ continuous.
Now we define the faces of $\phi$ for $0 \le i \le n$ by $$(\partial_i\phi)(s_0, \dots, s_n)=\phi(s_0, s_1, \dots, s_{i-1},0,s_i, \dots, s_n),$$ note that these are all singular $n-$simplices. Now for $i\not=j$ such that $(-1)^i=(-1)^j$ (which exists since $n+1 \ge 2$ by assumption and $0 \le i,j \le n+1$) we can define a homotopy between the singular $n-$simplices $(\partial_i \phi)$ and $(\partial_j \phi)$ as follows: $$ t\in[0,1], \quad \tilde{\phi}_{i \to j} = \begin{cases} \phi(s_0, s_1, \dots, s_{i-1}, ts_j, s_i, s_{i+1}, \dots, s_{j-1}, (1-t)s_i, s_j, s_{j+1}, \dots, s_n) & i<j \\ \phi(s_0, s_1, \dots, s_{j-1}, (1-t) s_i, s_j, s_{j+1}, \dots, s_{i-1}, ts_j, s_i, s_{i+1}, \dots, s_n) & j <i \end{cases} $$ Letting $(-1)^i = \varepsilon = (-1)^j$, we have that $\varepsilon \tilde{\phi}_{i \to j}(0) = \varepsilon \phi_i$ with $\phi_i(\Delta_n) \subseteq \bigcup_{i=0}^{n+1} (\partial_i \phi)(\Delta_n)$ and $\varepsilon \tilde{\phi}_{i \to j}(1) = \varepsilon \phi_j$ with $\phi_j(\Delta_n) \subseteq \bigcup_{i=0}^{n+1} (\partial_i \phi)(\Delta_n)$, thus the singular $n-$boundary "encloses" a homotopy between the singular $n-$simplices $\varepsilon \phi_i$ and $\varepsilon \phi_j$. This completes the proof. $\square$
Now since I am not proving that coordinate restrictions of $\phi$ are again continuous functions into $X$, I am not sure if this proof is obviously incorrect or obviously correct -- either way I'm sure it is a dumb question. In any case, if anyone who is an expert in this area wouldn't mind giving this a quick glance-over and me a sanity check, I would be much obliged for their help.
Context/Motivation for Question: So far I have been trying to use the following heuristic interpretation of (singular) homology to motivate the definitions and computations:
Specifically, for each dimension $n$ (and to motivate the definition of boundaries and why we factor them out from homology groups):
If you have any comments about the sensibility/dangerousness of these statements, please let me know, although it is not the main thrust of my question.