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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 $\sigma|_{[v_k,\dots,v_{k+l}]}=\sigma_2$?

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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.

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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.

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