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$X/\{0\}$ is the set of singletons of $X$ so we define a mapping $T(x) = x$. Since $\{0\} \subseteq \ker T$ we have a linear mapping $\hat{T}:X/\{0\}\to X$.

But $\{0\}$ is exactly equal to $\ker T$ so $X/\ker T \cong ran T = X$ since the mapping is surjective.

Does this make sense?

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    Yes, it does. And it is correct. – 2017-02-13
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    @freakish you're right. If you could make a brief answer to that effect, it would help resolve this post. – 2017-02-13
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    Minor nitpick: I think you should explicitly say that $T:X\to X$. That is, "we define a mapping $T:X\to X$ by $T(x)=x$." – 2017-02-13

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So yes, your reasoning seems to be correct. Although it took me a while to fully understand what you are doing here. So let's make it rigorous (for example by giving explicitely domains and codomains of each map).

Let $X$ be a group/ring/module/linear space (or anything that has an analogue of the first isomorphism theorem). Define

$$\mbox{id}:X\to X$$ $$\mbox{id}(x)=x$$

The famous identity map. Now this map is a homomorphism. Therefore by using the first isomorphism theorem we obtain an isomorphism

$$\overline{\mbox{id}}:X/\ker(\mbox{id})\to\mbox{id}(X)$$ $$\overline{\mbox{id}}(x\ker(\mbox{id}))=x$$

Since $\ker(\mbox{id})=\{0\}$ and $\mbox{id}(X)=X$ then this gives us an isomorphism

$$\overline{\mbox{id}}:X/\{0\}\to X$$

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    Thanks, I was using the same theorem but forgot it's name. – 2017-02-13