If $\mathbb{T}^2=\mathbb{S}^1\times\mathbb{S}^1$, let $S_2:=\mathbb{T}^2$#$\mathbb{T}^2$, where # denotes the connected sum.
I'm trying to find an intuitive explanation for the fact that $H^1(S_2)=\mathbb{R}^4$ (de Rham cohomology)
I've read somewhere that an intuitive way of understanding $H^1(M)$ was to think of how many ways you can strangle $M$ with a wire (here, "strangle with a wire" means finding a closed loop on $M$ which cannot be continuously shrunk to a point)
For example, there is no way to strangle $\mathbb{S}^2$, so that is why $H^1(\mathbb{S}^2)=0$. A cylinder $C=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=1\}$ can be strangled by the loop $\{x^2+y^2=1, z=0\}$ (or similar ones), so that is why $H^1(C)=\mathbb{R}$.
For the torus, you can find $2$ kinds strangling loops (I'll call them "big radius" and "small radius"):
which is why $H^1(\mathbb{T}^2)=\mathbb{R}^2$.
Now here is my problem with $S_2$. For each of the two holes in $S_2$, I can find $2$ strangling loops (big radius, small radius) just as in $\mathbb{T}^2$, which makes $4$ strangling loops. But what about this one in the junction:
Isn't it different from the other $4$? I though it was. If we take, for example, the strangling loop with small radius of the first hole, I would not be able to make it into the red one in the junction. Same with any of the other four.
Am I missing something? Or this is an example where this intuition just fails?
In fact the junction loop in your picture is the sum of all of the four other loops. I don't know how to make pictures, but your last picture is good. Look at it: it consists of an outside loop, and three inside loops with two points of intersection of the loops. Ok, now imagine that these three inside loops are just one single big loop, going from left back, to the central top, to right back, to right front, to central bottom (which you have dashed), to left front. We allow ourselves to ignore the points of contact. Viewed in this way the single big central loop is the same as the outside loop.
Hope that makes sense.