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Consider the universal covering space of $S^1\vee S^1$ which is well known as the direct limit of spaces like https://en.m.wikipedia.org/wiki/Rose_(topology)#/media/File%3ACayley_graph_of_F2.svg. However, one could also say that is is the Cayley graph of the free group generated by $a,b$.

My question is how to show that its fundamental group is trivial. Of course it is obvious from the picture and by the fact that it's a universal cover. However, I would like to compute that preferably using the limit construction. Can you help me? Thanks!

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    It's contractible. So its fundamental group is trivial. – 2017-01-24
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    Is see that every component of the direct limit (the picture shows one) is contractible. Does this property than also hold for the direct limit then? – 2017-01-24

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Here's an expansion of my comment.

You can prove contractibility of the whole space. The key property is that your direct limit space $T$, like its finite approximants, is a tree, meaning a space in which any two points are endpoints of a unique subspace homeomorphic to $[0,1]$. You prove the tree property for each finite approximant by induction, and then you prove the tree property for their union $T$. Then you can prove generally that any graph which is a tree is contractible.

Alternatively, once you've proved the tree property for each finite approximant, you don't have to go on to prove the tree property for their union $T$. Instead, you can use that for every compact space $K$ and every continuous function $\gamma : K \to T$, the image $\gamma(K)$ is contained in one of the finite approximants. Hence, if $K=[0,1]$ and $\gamma$ is a closed path then $\gamma$ is path homotopic to a constant in that finite approximant, which is a tree. Each $\gamma$ is therefore path homotopic to a constant in the the whole space $T$, and so $\pi_1(T)$ is trivial.