$\operatorname{Lk}(C_1,C_2)= \operatorname{Lk}(C_2,C_1)$ 
Let $D$ be an oriented link diagram with two components $C_1,C_2$. Then $\operatorname{Lk}(C_1,C_2)= \operatorname{Lk}(C_2,C_1)$.

By definition of Linking number is the following:

The linking number of $\operatorname{Lk}(C_1,C_2)$ is defined to be $[C_2] = \operatorname{Lk}(C_1,C_2)\cdot\mu_1\in H_1(S^3\setminus C_1)\simeq\Bbb Z=\langle\mu_1\rangle$.

I also have a pictorial definition (maybe same as wikipedia says). From that pictorial definition, the above statement is obvious. But from the definition using homology, it's not very obvious to me. How can I show they are the same numbers? * I know the concept of meridian
Note. I'm not very familiar with knot theory. I only know homology theory (Hatcher AT Chapter 2 level). That linking number definition is the first nontrivial definition I met.
 A: In my other answer, I use (co)homology and dualities to show this definition of linking number is symmetric - it's something I had worked out before and this was a good opportunity to write it down.
However, the question really only was "why is the homological definition the same as the diagram-based definition."  Here's how I've thought about it.  We're going to take the image of $C_2$ in $H_1(S^3\setminus C_1)$.
First, take a diagram of $C_1$ and $C_2$. Second, "lift" the $C_2$ part above the plane of the diagram, where when there's a part of $C_2$ that lies under $C_1$, add in line segments to detour under $C_1$ (this is illustrated below).  The lifted curve is homologous to the original curve because it's homotopic, but it's also easy to work out what boundaries of 2-chains you're adding in to make this happen.  Third, at the places where $C_2$ still goes under $C_1$, add in the boundary of a 2-chain to "bubble off" a loop around $C_1$.

This shows the 2-chain whose boundary, when added in, bubbles off the loop:

After bubbling off the loops, the remaining part of $C_2$ above the diagram lies in a subspace homeomorphic to an open ball $B$.  Recall that $H_1(B)=0$, so this means this part of $C_2$ is nullhomologous in $B$ (i.e., it is the boundary of a 2-chain in $B$), so it is the boundary of a 2-chain in $S^3\setminus C_1$ and can be removed.
All that's left are these rings on $C_1$.  If you are familiar with those "bead maze" toys for babies, the principle is now very similar: take all of the loops ("beads") and drag them along $C_1$ until they are co-located.  Unlike beads in a bead maze, loops have an orientation.  If two of the loops have opposite orientation, they cancel out.  What's left after cancellation is some multiple of $\mu_1$, and if you think through what we did, the coefficient in front of $\mu_1$ is the number of positive crossings where $C_2$ goes under $C_1$ minus the number of negative crossings where $C_2$ goes under $C_1$.  That matches Wikipedia's formula $\operatorname{Lk}(C_1,C_2)=n_1-n_4$ for linking number!

Side note: I'll take a moment to illustrate what the relation is between this and Alexander duality in the other answer.  If you lift up $C_2$ (letting it pass through $C_1$!) up above the plane of the diagram, then it is nullhomologous using a 2-chain in the ball $B$ that contains it.  Adding this to the 2-chains involved in the lift, we get a 2-chain $\Sigma_2$ in $S^3$ whose boundary is $C_2$.  Notice that with this setup, $C_1$ intersects $\Sigma_2$ exactly where we'd be bubbling off loops:

Pretty much the entire idea is that to account for the bubbled off loops, you can instead calculate the algebraic intersection number between $C_1$ and $\Sigma_2$.  Algebraic intersection number in this exemplary situation is from taking an oriented basis of tangent vectors of $\Sigma_2$ at the intersection along with the tangent vector of $C_1$ at the intersection, then taking the sign of the determinant of the $3\times 3$ matrix of these vectors.  This number tells you whether it's a $\mu_1$ or a $-\mu_1$ that bubbles off.
