$\Delta$-complex structure on the singular complex $S(X)$ I am confused with the definition of $S(X)$ so that I can't see why $S(X)$ is a $\Delta$-complex.

Here are some material by Allen Hatcher.

Though singular homology looks so much more general than simplicial homology, it can actually be regarded as a special case of simplicial homology by means of the following construction. For an arbitrary space $X$, define the singular complex $S(X)$ to be the $\Delta$-complex with one $n$-simplex $\Delta_\sigma^n$ for each singular $n$-simplex $\sigma:\Delta^n\to X$, with $\Delta_\sigma^n$ attached in the obvious way to the $(n-1)$-simplices of $S(X)$ that are the restrictions of $\sigma$ to the various $(n-1)$-simplices in $\partial\Delta^n$. It is clear from the definitions that $H_n^\Delta(S(X))$ is identical with $H_n(X)$ for all $n$.

And here it is the one of the three conditions of the definition of $\Delta$-complex:

A $\Delta$-complex structure on a space $X$ is a collection of maps $\sigma_\alpha:\Delta^n\to X$, with $n$ depending on the index $\alpha$, such that (i) The restriction $\sigma_\alpha\big|\mathring{\Delta}^n$ is injective, and each point of $X$ is in the image of exactly one such restriction $\sigma_\alpha\big|\mathring{\Delta}^n$.

Considering there is so much singular $n$-simplex, how does the condition holds?
In particular, why don't we may loss the injective condition?
If I get the idea of $S(X)$, then I will see that $H_n^\Delta(S(X))$ is identical with $H_n(X)$ for all $n$.
 A: I am thinking that perhaps your question is about the formalities of forming the quotient space $S(X)$ by gluing together simplices. So I'll write an answer along those lines.
To allay any possible confusion, I am going to imagine that the second yellow box has been rewritten with $X$ replaced by $Y$.
The goal is to construct the $\Delta$-complex 
$$Y=S(X)
$$
starting with any given topological space $X$. 
The $\Delta$-complex $Y$ is constructed as a certain quotient space. To do this carefully and rigorously, one needs a domain of the quotient map. I'll explicitly construct this domain, denoted $D$. 
Informally, $D$ is a disjoint union of copies of standard simplices, one for each singular simplex in $Y$. To do this formally, I'll need notation for the set of singular simplices.
Let $\Sigma$ be the set of all singular simplices $\sigma : \Delta^n \to X$, for all $n=0,1,2,...$. Let me break $\Sigma$ into disjoint subsets, one for each value of $n=0,1,2,...$, where $\Sigma_n$ is the set of all $n$-dimensional singular simplices $\sigma : \Delta^n \to X$. The domain of the quotient map is a disjoint union of copies of standard simplices, one for each element of the set $\Sigma$, using the $n$-dimensional standard simplex for elements of the subset $\Sigma_n$:
$$D = \sqcup_{n=0}^\infty \,\, \Sigma_n \times \Delta^n
$$
Using Cartesian products in this manner is a standard formal way to produce disjoint unions, by employing the index set with the discrete topology as one of the Cartesian factors (in this case, $\Sigma_n$).
Now we have to describe the gluing maps used to form $Y$ as a quotient space of $D$. There is one such gluing map for each $n$, each $\sigma \in \Sigma_n$, and each $n-1$ dimensional face $K \subset \Delta^n$. Namely, letting $i_K : \Delta^{n-1} \to K \subset \Delta^n$ denote the standard face map, and letting $\sigma | K \in \Sigma_{n-1}$ denote the composition 
$$\sigma | K = \Delta^{n-1} \rightarrow^{i_K} K \rightarrow^{\subset} \Delta^n \rightarrow^{\sigma} X
$$
for each $x \in \Delta^{n-1}$ we identity the point $(\sigma|K,x) \in \Sigma_{n-1} \times \Delta^{n-1}$ with the point $(\sigma,i_K(x)) \in \Sigma_n \times \Delta^n$. What this does is to identify the $n-1$ dimensional simplex 
$$\{\sigma | K\} \times \Delta^{n-1} \subset \Sigma_{n-1} \times \Delta^{n-1} \subset D
$$
with the $n-1$ dimensional face
$$\{\sigma\} \times K \subset \Sigma_n \times \Delta^n \subset D
$$
of the $n$ dimensional simple
$$\{\sigma\} \times \Delta^n \subset \Sigma_n \times \Delta^n \subset D
$$
Make all those identifications on $D$, one for each $n$, $K$, and $\sigma$, and you get the quotient $\Delta$-complex $Y=S(X)$.
A: It doesn't make sense to say that we might not have $\Delta$-complex structure because the space is big, because the space by construction is a $\Delta$-complex.
Given a topological space $X$, the singular complex $S(X)$ is defined to be the quotient space $\bigsqcup \Delta^k/\sim$ where $\Delta^k$'s are in bijective correspondence with the singular $k$-simplices $\sigma: \Delta^k \to X$ in $X$, and $\sim$ is defined by identifying a face $K$ of an $n$-simplex $\Delta^n$ corresponding to a singular $n$-simplex $\sigma$ in $X$ to the $(n-1)$-simplex corresponding to the singular $(n-1)$-simplex $\sigma|_K$, and the identification is done through a linear homeomorphism.
$S(X)$ is a $\Delta$-complex by construction: $f_\alpha : \Delta^n \to X$ are precisely the maps $\Delta^n \hookrightarrow \bigsqcup \Delta^k \to X$ where the first map is inclusion into some $n$-simplex and the second is the quotient map. As $\sim$ leaves the interior of each simplex alone, $f_\alpha$ is obviously injective on the interior of the simplices. 

Finally, it's not hard to see why simplicial homology of $S(X)$ coincide with the singular homology of $X$: the simplicial chain groups of $S(X)$ are isomorphic to the singular chain groups of $X$ because of bijection at the level of simplices and the boundary maps are exactly same because the simplices in $S(X)$ are identified according to the coincidences occurring to the singular simplices on $X$. 
(It is in fact true that not only the homologies coincide, $S(X)$ is weak homotopy equivalent to $X$; explicitly, the map $S(X) \to X$ given by sending each simplex in $S(X)$ to $X$ by the corresponding singular simplices is an isomorphism on homotopy groups. So for reasonably nice $X$, say, a CW-complex, they are homotopy equivalent)
