# Constructing a singular simplex from other two singular simplexes.

Suppose $$\sigma_1:\Delta^k \rightarrow X$$ is a singular $$k$$-simplex and $$\sigma_2:\Delta^l \rightarrow X$$ is a singular $$l$$-simplex. Is there a singular $$(k+l)$$-simplex, $$\sigma: \Delta^{k+l} \rightarrow X$$, such that $$\sigma|_{[v_1,\dots,v_k]} = \sigma_1$$ and $$\sigma|_{[v_k,\dots,v_{k+l}]}=\sigma_2$$?

Were there a common extension $$\sigma : \Delta^2 \to X$$ with $$\sigma|_{[v_0, v_1]} = \sigma_1$$ and $$\sigma|_{[v_1, v_2]} = \sigma_2$$ where 1-simplices $$\sigma_1$$ and $$\sigma_2$$, then it must be the case that $$\sigma_1(v_1) = \sigma_2(v_1)$$. This won't always be possible: consider the case when the $$\sigma_1$$ and $$\sigma_2$$ have disjoint images.
Well, since you want a singular simplex you only require the mapping $$\sigma: \Delta^{k+l}\to X$$ to be continuous. So you can certainly, given $$\sigma_1$$ and $$\sigma_2$$, identify $$k$$ and $$l$$ faces of $$\Delta^{k+l}$$ corresponding to the simplices above, and just arbitrarily define how $$\sigma$$ maps the other faces (in a continuous way, of course).
For example, suppose that you have two singular $$1$$-simplices $$\alpha$$ and $$\beta$$. Then you can identify two of the sides of a singular $$2$$-simplex with $$\alpha$$ and $$\beta$$, but there is a third face which you may define arbitrarily.